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Chapter  9. Infinite Sequences and Infinite Series

Calculus

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Chapter  9. Infinite Sequences and Infinite Series

Section 9.1 Sequences and Series

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A list of numbers in a definite order, say, , can be given precise mathematical meaning. A sequence is a function whose domain is the set of positive integers. If , for the number , it is customary to denote the sequence by the notation or simply or  sometimes by : . The range of , that is, the values , are called the terms of the sequence. The number is called the first term of the sequence, the second term, and so on, and , the th term.

An example is the sequence , in which , or writing out its terms,

.

Another example is the sequence , in which or written in detail,

.

Notice that does not have to start at

Sequences are frequently specified by simply giving a formula for the term. For example, if  , then the sequence is

.

Another way to specify a sequence is by means of a recursion formula. As an example, consider the sequence called the Fibonacci sequence whose terms satisfy the following conditions:

and

Written in detail, the sequence is

.

Decimal Expansion

The decimal expansion of a real number is its representation with base-10 (That is, the decimal system). For example, the decimal expansion of is 625, of is 3.14159, and of is 0.1111.

The decimal expansion of a real number can be found in Sage using the following commands: R = RealField(200); R(17/6); R(pi); R(e).

The decimal expansion of a number may terminate (in which case the number is called a regular number or finite decimal, for example, ), eventually become periodic (in which case the number is called a repeating decimal, for example, ), or continue infinitely without repeating (in which case the number is called irrational).

The following table summarizes the decimal expansions of the first few unit fractions. As usual, the repeating portion of a decimal expansion is conventionally denoted as given in the following table.

 fraction decimal expansion fraction decimal expansion 1 1 1/11 1/2 0.5 1/12 1/3 1/13 1/4 0.25 1/14 1/5 0.2 1/15 1/6 1/16 1/7 1/17 1/8 1/18 1/9 1/19 1/10 0.1 1/20

Limit of a Sequence

Suppose that is a given sequence. A real number is said to be the limit of the sequence , written as

or, alternatively, as , if the terms can be made as close to as possible by taking to be sufficiently large. The symbol is read “ tends to infinity.”

It is clear that if has a limit, then .

Figure 1 illustrates the definition of a sequence by showing the graphs of two sequences that have the limit .

The intuitive definition of limit of a sequence given above can be made more precise. We say that a sequence has the limit and we write

(a)                                        (b)

(c)                                       (d)

Figure 1 Graphs of two sequences with

or, alternatively, as , if for any given , there is a corresponding integer (which may depend on ) such that

whenever .

If exists, we say that the sequence converges (or is convergent). Otherwise, the sequence is said to diverge (or is divergent).

Remember that represents the distance between and . The inequality is equivalent to the double inequality . This definition can, therefore, also be stated as follows: means that whatever interval is taken (with ), all the terms of the sequence except possibly the first (depending on )  lie within this interval.

The sequence is said to diverge (or diverges to )  and we write

if, for any given positive number , there is an integer (depending on ) such that whenever , in other words, becomes large as becomes large. Similarly, if for any given positive number , there is an integer (depending on ) such that   whenever , we write

.

Archimedean Property : Given any real number , there exists a natural number such that .

EXAMPLE 1

Prove that

Solution.  Consider a given We have to determine (depending on ) so that

for every

Now if Let be any integer larger than (follows from Archimedean  Property). Then we hav

if

as desired.

EXAMPLE 2

Prove that

Solution. Consider a given   We have to determine (depending on ) so that

for every

Now if . Let be any integer larger than (follows from Archimedean Property). Then we have

if

as desired.

EXAMPLE 3

Prove that if , then

Solution.  Let be given. We have to determine (depending on ) such that

for every

We note that if , then for some . By the binomial theorem,

and therefore

if .

Let be any integer larger than . Then we have

if .

This completes the proof.

EXAMPLE 4

Prove that if , then

.

Solution.  Let be given. We have to determine (depending on ) such that

for every

Since , we have for some . By the binomial theorem,

,

and therefore

if .

Let be any integer larger than . Then we have

if .

This completes the proof.

The  following  theorem  immediately follows  from  Example 2  and  Example  3.

THEOREM 1  Limit of a geometric sequence

If the sequence converges with

and  diverges for all other values of (See Figure 2.)

Figure 2

The definition of the limit of a sequence is very similar to that of a limit of a function at infinity. The only difference between and is that must be an integer. This gives rise to the following result, which proves to be useful for the evaluation of limits of sequences.

THEOREM 2

If and when is an integer, then . (See Figure 3.)

Figure 3

The fact that  if , can also be adapted from Theorem 1 and also holds .

The following limit laws of sequences, which are very similar to the limit laws of functions, can be deduced from Theorem 1 and are easily proved.

THEOREM 3  Limit Laws for Sequences

If and are convergent sequences and is a constant, then

1.

2.

3. for any constant

4.

5. if

6. if and

The Squeeze Theorem can be adapted for sequences.

THEOREM 4  Squeeze (Sandwich) Theorem for Sequences

If for and , then

.

Another useful result follows upon the application of the Squeeze Theorem for sequences.

If , then .

EXAMPLE 5

Find  .

Solution. Dividing numerator and denominator by the highest power of in the denominator, and then using the Limit Laws, we have

.

EXAMPLE 6

Find  .

Solution. Even though both numerator and denominator approach infinity as L’Hospital’s Rule does not apply directly because numerator and denominator are not continuous. However, if we consider the related differentiable function , then we can apply L’Hospital’s Rule. In this case

.

Consequently,

.

EXAMPLE 7

The sequence diverges because its terms oscillate between and so that they do not approach any number as .

EXAMPLE 8

Evaluate if it exists.

Solution. The limit of the absolute value is found to be

Therefore, by the application of the Squeeze Theorem for sequences

.

Criterion for Convergence of Monotonic Sequences

We say that a sequence is

(a) strictly increasing if for all ;

(b) strictlydecreasing if for all .

A sequence is called monotonic if it is either increasing or decreasing.

EXAMPLE 9

Prove that the sequence is decreasing.

Solution.

1. We must show that , that is,

By cross‐multiplication, we get the numerator of :

.

Since for all , . Hence, for all .       Therefore, is decreasing.

2. Let

Consider the related function . Since

when

is decreasing on and so for all .

Therefore, is decreasing.

One may find a positive number such that the inequality  or is satisfied for every , however large. In such a case the sequence is said to be bounded. For example, is a bounded sequence, but is not. The following theorem says that every convergent sequence is bounded.

THEOREM 5

Any convergent sequence is bounded.

Proof. Suppose that Taking in the definition of limit, there exists an integer such that if

We have for Therefore, if we define  we have for every

The converse of Theorem 5 is not true: there exists a sequence which is bounded but not convergent, for example, the sequence is one such example. Thus not every bounded sequence is convergent. Also, not every monotonic sequence is convergent. For instance, consider . However, if a sequence is both bounded and monotonic, it must be convergent. (See Figure 4.)

THEOREM 6  Monotonic Sequence Theorem

Every bounded, monotonic sequence is convergent.

(a) increasing and bounded above       (b) decreasing and bounded below

Figure 4

EXAMPLE 10

Determine  whether  the sequence defined  by  the recurrence relation

for

is convergent or divergent. If so, find the limit of the sequence.

Solution. We first observe the first several terms:

.

The sequence and its terms seem to be increasing and approach 2. Mathematical induction will be used to establish for all . This is true for because Assuming that it is true for ,  we have

so and

which implies that is also true for . Therefore the inequality is true for all by induction.

We next prove that is bounded. Induction will also be used to verify that for all . This is true for since  . Supposing that it is true for , we have

and

so that is also true for . By induction, for all . Note that for all . Thus, for all .

Since is increasing and bounded, and exists by the Monotonic Sequence Theorem.1) Moreover, the limit must satisfy the relation whence it follows that or .

Thus since all terms are positive.

CAS EXAMPLE 11

Determine whether the sequence converges or diverges. If it converges, find the limit.

Solution.

var('x i n')

p1=plot((3 -2*x)/(1 +5*x), (x, 1, 15), rgbcolor=(1,0,0))

p2=list_plot([(3 -2*x)/(1 + 5*x) for i in range(0, 16, 1)], rgbcolor=(0, 0, 1))

show(p1+p2)

limit((1 - 2*n^2)/(2 + 5*n^2), n=+oo)

Figure 5

CAS EXAMPLE 12

Determine whether the sequence converges or diverges. If it converges, find the limit. .

Solution.

var('n')

limit((3 -  1/(3^n))* (2 +  1/(2^n)), n=+oo)

Series

Given a sequence of numbers, one might be led to add up the infinite number of its terms. Defining and implementing this summability in a mathematically precise sense requires the use of the limit.

Given a sequence we use the symbol

or, more concisely when there is no possible ambiguity,

.

This symbol is called an infinite series, or just a series. With we associate a new sequence , where

.

The numbers are called the partial sums of the series If the sequence converges to , we say that the series converges (or is convergent), and write

.

The number is called the sum of the series in other words, by definition,

provided the limit exists. It should be understood that is the limit of a sequence of partial sums, and is not simply obtained by addition.

If diverges, the series is said to diverge (or is divergent).

For example, for the series

the partial sums are as follows: 1, 3, 6, 10, 15, 21, . After the th term we get , which becomes very large as increases. Thus this series diverges. On the other hand, for the series

the sequence of its partial sums is given by

,

which converges to 1. Therefore, the series converges and its sum is

Let and be two sequences such that  , .  In this case is called a telescoping series. In mathematics, a telescoping series2) is a series whose partial sums eventually only have a fixed number of terms after cancellation. Such a technique is also known as the method of differences. The approach used to find a formula for the partial sums is to find the sum of the first terms and then take the limit as approaches infinity:

EXAMPLE 13

Show that the series is convergent, and find its sum.

Solution.

Using the partial fraction decomposition

〔Telescoping series〕

the th partial sum can be written as

and hence,

Therefore, the series converges and its value is

The simplest of all series is perhaps the geometric series .

THEOREM 7  Geometric Series

If the series converges, and

If the series diverges.

Proof.

If the partial sums are

If and hence, the series converges and its sum is

If then the series evidently diverges. For , so  diverges and for diverges and hence, does not exist.

EXAMPLE 14

Find the sum of the geometric series

Solution.  The series is a geometric series with the first term and the common ratio . Since , the series converges and its sum is .

EXAMPLE 15

Determine whether the series is convergent or divergent.

Solution.  The series can be rewritten in a form that we recognize.

.

This is a geometric series with and Since the series diverges.

EXAMPLE 16

Write the number as a ratio of integers.

Solution.  We have

.

After the first term, the right side is a geometric series with and . Therefore,

.

THEOREM 8  Necessary Condition for Convergence

If the series converges, then .

Proof. Note that . If the series converges to then and . Thus,

Figure 6 shows if converges.

Figure 6

The condition is not, however, sufficient to ensure convergence of . That is, a series may diverge even though A classical example illustrating this case is the series called the harmonic series

.

This series diverges; for proof we refer to Theorem 2 in Section 9.2.

The contrapositive of Theorem 8 provides a useful test for divergence.

A Test for Divergence

If does not exist or if , then the series is divergent.

EXAMPLE 17

Prove that the series diverges.

Solution.  Since , the series diverges by the test for divergence.

We know that the sum of a convergent series is the limit of its partial sums. Using the properties of limits of sequences, the following properties of convergence can be easily established.

THEOREM 9  Sum of Two Convergent Series

Let be a constant. If and are convergent series, then the series , , and are also convergent and we have

(ⅰ) ,

(ⅱ) ,

(ⅲ)

Thus two convergent series may be added or subtracted term by term, and the resulting series converges to the sum or difference of the two series.

EXAMPLE 18

Find the sum of the series .

Solution.

From Example 13 we have

.

The second series is a geometric series with and , and its sum is

.

Therefore, the series converges by Theorem 9 and it's sum is

.

var('n')

a(n) =  3/((n+1)*(n+2))

P=sum(a(n), n, 1, +oo)

b(n) =  2/(3^n)

Q=sum(b(n), n, 1, +oo)

P-Q

Remark  The convergence or divergence of a series is not affected by adding to or subtracting from it a finite number of terms. This can be inferred from the relation

.

If the series converges, then the full series

also converges. A similar argument applies if the series diverges.

9.1 EXERCISES (Sequences and Series)

1-3. Find a formula for the general term of the sequence, assuming that the pattern of the first few terms continues.

1.

Solution. .

Therefore, .

2.

Solution. .

Therefore, .

3.

Solution.

Therefore, .\

4-15. Determine whether the sequence converges or diverges. If it converges, find the limit.

Solution.

var('x, i, n')

p1 = plot((1-x^3)/(2+3*x^2), (x, 1, 15), rgbcolor=(1,0,0))

p2 = list_plot([(1 - i^2)/(2 + 3*i^2) for i in range(0, 16, 1)], rgbcolor=(0,0,1))

show(p1+p2)

limit((1 - n^3)/(2 + 3*n^2), n=+oo)

Solution.

var('x, i, n')

p1 = plot((x^2 - 2*x + 3)/(2*x^3 + 2), (x, 1, 50), rgbcolor=(1,0,0))

p2 = list_plot([(i^2 - 2*i + 3)/(2*i^3 + 2) for i in range(0, 51, 1)], rgbcolor=(0,0,1))

show(p1+p2)

var('n')

limit((n^2 - 2*n + 3)/(2*n^3 + 2), n=+oo)

CAS 6.

Solution.

var('a, b, i, n')

p1=plot(a^2/(3 + 2*a^2), (a, 1, 20), rgbcolor=(1,0,0))

p2=plot((-1)*b^2/(3 + 2*b^2), (b, 1, 20), rgbcolor=(1,0,0))

p3=list_plot([((-1))^i*i^2/(3 + 2*i^2) for i in range(0, 21, 1)], rgbcolor=(0,0,1))

show(p1+p2+p3)

var('n')

limit(((-1))^n*n^2/(3 + 2*n^2), n=+oo)limit(ln(3*n)/ln(5*n), n=+oo)

CAS 7.

Solution.

var('x, i, n')

p1=plot((sin(x))^2/(3^x), (x, 1, 20), rgbcolor=(1,0,0))

p2=list_plot([(i,(sin(i))^2/(3^i)) for i in range(1,21,1)], rgbcolor=(0,0,1))

show(p1+p2)

limit((sin(n))^2/(3^n), n=+oo)

CAS 8.

Solution.

var('x, i, n')

p1=plot(ln(3*(x))/ln(5*(x)), (x, 1, 40), rgbcolor=(1,0,0))

p2=list_plot([(i, ln(3*i) /ln(5*i)) for i in range(1,41,1)], rgbcolor=(0,0,1))

show(p1+p2)

limit(ln(3*n)/ln(5*n), n=+oo)

CAS 9.

Solution.

var('x, i, n')

p1=plot((1- 1/x)^x, (x, 1, 20), rgbcolor=(1,0,0))

p2=list_plot([(i,(1- 1/i)^i) for i in range(1,21,1)], rgbcolor=(0,0,1))

show(p1+p2)

limit((1- 1/n)^n, n=+oo)

CAS 10.

Solution.

var('x, i, n')

p1=plot((3/x)^(2/x), (x, 1, 20), rgbcolor=(1,0,0))

p2=list_plot([(i,(3/i)^(2/i)) for i in range(1,21,1)], rgbcolor=(0,0,1))

show(p1+p2)

limit((3/n)^(2/n), n=+oo)

11.

Solution.

Here, and .

Thus, .

Hence, this sequence is divergent.

12.

Solution.

Thus,

Hence, this sequence is divergent.

13.

Solution.

Hence, this sequence is convergent.

14.

Solution.

By L’Hopital’s Rule,

Hence, this sequence is convergent.

15.

Solution.

Since , Therefore,

Hence, this sequence is convergent.

16. Investigate the sequence defined by the recurrence relation for . In particular, show that

.

Solution.

Therefore,

17. Let

(a) Determine whether is convergent.

(b) Determine whether is convergent.

Solution.

(a)

Therefore, is convergent.

(b)

Therefore, is divergent.

18-24. Determine whether the following series is convergent or divergent. Find the sum if it is convergent.

18.

Solution. .

Thus, is divergent.

var('n')

a(n) =  ((-1)^n *n)/(n+1)

sum(a(n), n, 1, +oo)

19.

Solution.

This is a geometric series and since

, this series is convergent.

And, .

var('n')

a(n) =  ((e)/10)^n

sum(a(n), n, 1, +oo)

20.

Solution.

But, is not convergent and is convergent.

Thus, is divergent.

21.

Solution.

Therefore, is convergent to

22.

Solution. By L’Hopital’s Rule,

Therefore, is divergent.

23.

Solution. Since , this geometric series with () is convergent.

And, .

24.

Solution.

As ,

Thus, is divergent.

25-26. Express the number as a ratio of integers.

25.

Solution.

26.

Solution.

var('n')

a(n) =  83/(10^(2*n))

sum(a(n), n, 1, +oo)

9.2 Tests for Convergence of Series

Finding the exact sum of a series is a very difficult task, except for series with simple forms such as for geometric series or telescoping series. It is often enough to know the convergence or divergence of a series. If we learn that a given series converges, we may obtain as close an estimate as we wish for its exact sum, which will often suffice for both theoretical and applied purposes. In the following sections, several tests will be introduced that enable us to determine whether a series converges or not.

Examples of series with known convergence may be obtained by applying the integral test due to Colin Maclaurin (1698-1746), which, as the name indicates, involves integration. As a motivating example, let us consider the series

.

By geometric observation and calculation it can be shown that

for all . (see Figure 1.) The sequence of partial sums is thus bounded above and is also increasing. Hence by the Monotonic Sequence Theorem, the  partial sums of the sequence converge. Therefore the series is convergent. Consider as another motivating example,

.

By geometric observation(see Figure 2) and calculation it can also be shown that

Figure 1                Figure 2

Since the right side of the inequality goes to as , we get . Hence is divergent.

It should be noted that the convergence and divergence of the series and were essentially determined by the convergence and divergence of the improper integrals and , respectively. Also note that the functions and , which are obtained from the -terms and by replacing with are continuous, positive and  decreasing functions on .

In general for series where the function () is continuous, positive and decreasing on , the following powerful test for convergence applies.

THEOREM 1 The Integral Test

Suppose is a continuous, positive, decreasing function on and let . Then the series is convergent if and only if the improper integral is convergent. In other words,

(i) If is convergent, then is convergent.

(ii) If is divergent, then is divergent.

Proof. The area of the first shaded rectangle in Figure 3 is the value of at the first right endpoint which is . So, by comparing the areas of the shaded rectangles with the area under from 1 to , we can write

[1]                            .

On the other hand, Figure 3 shows that

[2]

Figure 3 Subject to the conditions of the integral test, either the series and the integral both converge or diverge.

(i) If is convergent, then [1] gives

since . Therefore,

.

Since this is true for all , the sequence is bounded. Also is an increasing sequence since , due to . By the Monotonic Sequence Theorem, is convergent. Therefore, is convergent.

(ii) If is divergent, then as because . But [2] gives

and so, . This implies that and hence, diverges.

We shall show how the Integral Test can be applied to the -series .

THEOREM 2  ‐series Test

The series converges if and diverges if

Proof.  If , then the function is clearly continuous, positive, and decreasing on . We know that

converges if and diverges if .

By the Integral Test, the series converges if and diverges if .

For , we have . If , then . In either case the series diverges by the Test for Divergence.

As an immediate result of Theorem 2, we obtain the fact that the harmonic series diverges because it is a -series with

EXAMPLE 1

Test the series for convergence or divergence.

Solution.

The function is continuous, positive, and decreasing on and we have

.

Since the integral converges, by the Integral Test, the given series  is convergent.

EXAMPLE 2

Test the series for convergence or divergence.

Solution.

The function is positive and continuous for . Since is increasing, is decreasing for .

We have

,

which is divergent. The series is divergent by the Integral Test.

The next test shows that if the terms of a series with positive terms are dominated by the corresponding terms of a convergent series, then the first series is convergent. It also shows that if the terms of a series with positive terms dominate corresponding terms of a divergent series, then the first series is divergent.

THEOREM 3  The Comparison Test

Let and be series of positive terms with for all , where is some fixed integer.

(i) If  is convergent, then is also convergent.

(ii) If is divergent, then is also divergent.

Proof.  (i) Consider

,    ,    .

Since both series have positive terms, the sequences and are increasing . Also, , so for all . Since , we have . Thus, for all . Therefore is convergent by the Monotonic Sequence Theorem. Hence, is convergent.

(ii)If is divergent, then since is increasing. But , so . Hence as increases. Therefore, diverges

The comparison test is very useful and is most frequently used to prove whether a series converges or diverges. To use it effectively, we must have at our disposal examples of series of positive terms where convergence or divergence of the series is known. Two of the most important series for comparison purposes are the geometric series and the ‐series.

EXAMPLE 3

Test the series for convergence or divergence.

Solution. The dominant term in the denominator is for large . Note that

for

Since the -series converges, by the Comparison Test, the given series must also converge.

EXAMPLE 4

Use the Comparison Test to test for convergence or divergence of the series .

Solution. Since for , we .

The series is divergent by -series with .

The given series is divergent.

Note for the series , the inequality

for

is not useful for an application of the Comparison Test. In general, if the terms of the given series are larger than the terms of a convergent series or smaller than those of a divergent series, the Comparison Test may not be useful. In such cases the following test can be used.

THEOREM 4  The Limit Comparison Test

Let and be series of positive terms. Assume that  exists. Then

converges if and only if converges.

diverges if and only if diverges.

Proof.

Let and be positive numbers such that . Because is close to for large , there is an integer such that

when

and so,

when .

If converges, so does Thus, converges by part (i) of the Comparison Test. If diverges, so does , and hence, part (ii) of the Comparison Test shows that diverges.

EXAMPLE 5

Determine if the series is convergent or divergent.

Solution.

The th term behaves approximately like

, and let and

Since

and converges, by the Limit Comparison Test, the given series converges.

EXAMPLE 6

Test the series for convergence or divergence.

Solution.

The th term behaves approximately like . Let and .  Since

and diverges, by the Limit Comparison Test, the given series diverges.

CAS EXAMPLE 7

Test for convergence or divergence of the series

.

Solution.

var('n')

un=2^n*sin(pi/(2^n))

limit(un, n=+oo)            # pi

Answer : Since , the series diverges.

6.2 EXERCISES (Tests for Convergence of Series)

1-5. Determine whether the series is convergent or divergent using the Integral Test.

1.

Solution.

The function is continuous, positive, and decreasing on .

The Integral Test applies.

Hence by the Integral Test,  diverges.

CAS 2.

Solution.

var('n')

a(n) = n*e^(-n)

sum(a(n), n, 1, +oo)    #integral(a(n), n, 1, +oo)  # gamma(2, 1)

Answer : -e/(2*e - e^2 - 1)

Thus, the series converges.

3.

Solution.

is continuous and positive on , and also decreasing since for . Hence we can use the Integral Test.

Thus, converges.

And by the Comparison Test, is convergent.

var('n')

a(n) = exp(1/n^2)/n^2

sum(a(n), n, 1, +oo)

Answer : sum(e^(n^(-2))/n^2, n, 1, +Infinity)

4.

Solution. The function is continuous, positive on , and also decreasing. hence by Integral test, since for , so we can use the Integral Test.

Hence by Integral Test diverges.

5.

Solution. The function is continuous, positive on , and also decreasing, since if , so we can use the Integral Test and .

Hence by the Integral Test, converges.

6-7. Find the values of for which the series is convergent.

6.

Solution. When , is continuous and positive on , and also decreasing since    for

( for ). Hence we can use the Integral Test.

Thus, when , diverges.

When , is continuous and positive on , and also decreasing since

for , so we can use the Integral Test.

This limit exists whenever , so the series converges for .

7.

Solution.

We have already shown (in Exercise 4) that when , the given series is divergent. Let us assume .

The function is continuous, positive on , and also decreasing since

if ,

so we can use the Integral Test and .

Thus the series converges for .

8. Find all positive values of for which the series converges.

Solution.

The function is continuous, positive on , and also decreasing since

if , ,

so we can use the Integral Test:

Thus for. The series converges for .

9-13. Test for convergence or divergence of the series.

9.

Solution. , so the series converges by comparison with the -series .

10.

Solution. Use the Limit Comparison Test with and .

Since converges,  converges.

CAS 11.

Solution.

var('n')

u(n)=2^n*factorial(n)/n^n

limit(u(n+1)/u(n), n=+oo)  #   2*e^(-1)

bool(2*e^(-1)<1)             #   True

CAS 12.

Solution.

var('n')

u(n)=n^2/3^n

limit(u(n+1)/u(n), n=+oo)   #   1/3

13.

Solution. Use the Limit Comparison Test with

and .

Then and are series with positive terms and

.

Since converges, converges.

14. If is a convergent series with non-negative terms, is it true that is also convergent?

Solution. Use the Comparison Test.

If , for all , then is convergent.

But, if , for all then we cannot decide the convergence of .

15. If is a convergent series with positive terms, is it true that is also convergent?

Solution.

Yes. Since is a convergent series with positive terms, , and is a series with positive terms (for large enough ).

We have .

Thus, is also convergent by the Limit Comparison Test.

9.3 Alternating Series and Absolute Convergence

Alternating Series

In this section we will focus on tests for convergence for series with terms that are not necessarily positive.

There is a special type of series with terms that are alternately positive and negative, , the are positive, it is called an alternating series.

The convergence of an alternating series is easily determined by the following result.

THEOREM 1  The Alternating Series Test

The alternating series  , where for all

.is convergent if it satisfies

(ⅰ) for all ,

(ⅱ)

Proof.

Let us first consider the partial sums of even orders

,

.

By hypothesis, the terms are decreasing, so that the quantity in the parenthesis is positive. Hence , so that the partial sums of even orders form an increasing sequence. On the other hand,

.

Figure 6 Alternating Series

Since the quantity in each parenthesis is positive, for all . Hence the sequence is increasing and bounded. By the Monotonic Sequence Theorem, it must converge to some value .

Now,

,

and

.

Since , so the partial sums of even and odd orders tend to the same limit .

Thus the series is convergent.

Theorem 1 enables one to easily establish the convergence of an alternating series, since all that is necessary is to show that both (ⅰ) and (ⅱ) hold as follows:

(ⅰ) for all ,

(ⅱ)

EXAMPLE 1

Test the convergence of the alternating harmonic series

.

Solution. The sequence where satisfies

(ⅰ)   because  ,

(ⅱ) .

By the Alternating Series Test, the series is convergent.

EXAMPLE 2

Determine whether the serie is convergent or divergent

.

Solution.

For the sequence where , we have

.

So condition (ii) is not satisfied.

Thus this series is divergent by the Alternating Series Test.

Absolute Convergence, Conditional Convergence, and the Ratio and Root Tests

DEFINITION2

A series is absolutely convergent  if is convergent.

A series is conditionally convergent if converges, but diverges.

For series with positive terms, there is no distinction between ordinary convergence and absolute convergence. We have the following relation.

THEOREM 3

If a series is absolutely convergent, then the series is convergent.

The series is absolutely convergent because

is a convergent ‐series . It is therefore convergent by Theorem 3.

The converse of Theorem 3 is false: a classical example is given by the alternating harmonic series

which converges by Theorem 1 (or see Example 1), but it does not converge absolutely because the corresponding series of absolute values

is divergent.

Absolute convergence is a stronger condition than ordinary convergence. A series is said to be  conditionally convergent if it is convergent but not absolutely convergent. Thus the series is conditionally convergent.

EXAMPLE 3

Is the series absolutely convergent?

Solution.

The first three terms are positive, the next three are negative, and the following three are positive. The signs change irregularly. This series has both positive and negative terms, but it is not alternating. series. So, apply the Comparison Test to the series of absolute values

.

Since for all , we have . Since is convergent (‐series with ), by the Comparison Test, is convergent. Thus, the given series is absolutely convergent and therefore convergent by Theorem 3.

The ratio test, see below is very useful when determining whether a given series is absolutely convergent.

THEOREM 4  The Ratio Test

(ⅰ)If , then the series is absolutely convergent (and therefore convergent).

(ⅱ)If or , then the series is divergent.

(ⅲ)If , the Ratio Test is inconclusive; that is, no

Proof.  (i)Since , there exists a number such that . Since

and

the ratio will eventually be less than ; that is, there exists an integer such that

whenever

or, equivalently,

[1]                           whenever .

Putting successively equal to in [1], we obtain

and, in general,

[2]                             for all  .

Now, the series

is convergent because it is a geometric series with . So, the inequality [2], together with the Comparison Test, shows that the series

is also convergent. It follows that the series is convergent (Recall that a finite number of terms do not affect convergence.). Therefore, is absolutely convergent.

(ii) If or , then the ratio will eventually be greater than 1; that is, there exists an integer such that

whenever .

This means that whenever and so .

Therefore, diverges by the Ratio Test.

EXAMPLE 4

Investigate the convergence of the series .

Solution. With , we have

as .

Since , the given series is absolutely convergent by the Ratio Test and therefore the series is convergent.

EXAMPLE 5

Investigate the convergence of the series .

Solution. Here , we have

Since , the given series diverges by the Ratio Test.

EXAMPLE 6

Investigate the convergence of the series .

Solution. Here .  Note that the series is convergent by the -series test. But the Ratio Test is inconclusive.

EXAMPLE 7

Investigate the convergence of the series .

Solution. Here . Then

and the Ratio Test is inconclusive. By using the -series test, the given series is shown to be divergent.

Examples 6 and 7 shows part (iii) of the Ratio Test: the condition does not allow one to make a conclusion on the  convergence or divergence of the series . In this case another test should be considered.

When th powers occur, it is convenient to apply the following test:

THEOREM 5  The Root Test

(ⅰ) If , then the series is absolutely  convergent (and therefore convergent).

(ⅱ) If  or , then the series is divergent.

(ⅲ) If , the Root Test is inconclusive.

EXAMPLE 8

Is the series convergent?

Solution. Let . Then

.

Since , the series converges by the Root Test.

EXAMPLE 9    This problem will be replaced by a Root Test problem soon)

Test for convergence or divergence, of the alternating series:

Solution.

var('n')

u(n)=1/log(n)

assume(n>2)

bool(u(n+1)<u(n))  #  True

limit(u(n), n=+oo)

Thus is convergent.

9.3 EXERCISES (Alternating Series and Absolute Convergence)

1-7. Test for convergence of the following alternating series:

1.

Solution.

Since converse to zero and is decreasing, this alternating series is convergent by the alternating series test.

2.

Solution. Note that then ,

.

By Alternating Series Test, the series is divergent.

3.

Solution.

It satisfies because and .

By the Alternating Series Test, the series is convergent.

CAS 4.

Solution. Note that

var('n')

u(n)=(-1)^(n+1)*n/(n+1)

limit(abs(u(n)), n=+oo)

5.

Solution.

Since converse to zero and is decreasing, this alternating series is convergent by the alternating series test.

6.

Solution. ,

By the Ratio Test, this series is absolutely convergent.

Hence, is convergent.

7.

Solution.

By Alternating Series Test, the series is divergent.

8. When , is the series convergent?

Solution. Let .

For ,

for all .

And for ,

Thus, by the Alternating Series Test, for , this series is convergent.

9-14. Test whether the series is absolutely convergent, conditionally convergent, or divergent.

9.

Solution. ,

.

Hence, the series is divergent.

CAS 10.

Solution. Divergent by the Ratio test.

var('n')

u(n)=exp(-n)*factorial(n)

limit(u(n+1)/u(n), n=+oo)

11.

Solution. .

Since for all , we have . Thus Let

is absolutely convergent.

By the Comparison Test, is absolutely convergent, and hence convergent.

12.

Solution. .

Using the root test,

is absolutely convergent, and hence convergent.

13.

Solution. ,

We do not know if this series is absolutely convergent, when using the ratio test. Let us try another test.

Here, for all and

By the Alternating Series Test,

is conditionally convergent.

14.

Solution.

Then

This means we cannot conclude convergence of this series using the Ratio Test.

Consider .

Here, for all and

By the Alternating Series Test, is conditionally convergent.

9.4 Power Series

The most important series are power series. Any series of the form

[1]

is called a power series with center or a power series about . Here, the

symbols and denote real numbers called the center and coefficients of the series, respectively and denotes a real variable. We are interested in criteria for the convergence of the power series ([1] is related to Taylor Series in Section 9.5). In general, the series [1] will converge or diverge, depending on the choice of . The power series [1] always converges when , adopting the convention that even when . In general, it may converge for some values of and diverge for other values of .

The geometric series

with the common ratio is a power series with center . We know that this series converges if and diverges if It is very surprising that every power series turns out to exhibit a behavior similar to the geometric series. More specifically, for every power series there is an associated interval, the interval of convergence for the series [1]. This is an interval such that [1] converges if is inside of the interval and diverges if is outside the interval. The behaviour at the endpoints of the interval of convergence varies, depending on the series.

THEOREM 1

For a given power series there are only three

possibilities:

(ⅰ) The series converges only when .

(ⅱ) The series converges for all .

(ⅲ) The series converges if , for some positive number , and diverges if  . ( one must check.)

Figure 1

DEFINITION2

For a given power series , the series converges if    for some positive number , and diverges if  . The positive number is called the radius of convergence of the power series. To cover all cases, we consider a point as an interval of (case (i)), and the set of real numbers as the inside of an interval of (case (ii)). The interval that consists of all values of for which the series converges is called the interval of convergence of the power series.

It should be noted that the case is not covered in Theorem 1.

The reason for this is that some series may converge at and others

diverge. For example, may diverge for , whereas converges for which can be shown by using the same techniques as those in the examples. Therefore, the convergence or divergence of the series at the endpoints of the interval, that is, at should be investigated separately. In general, the radius of convergence is determined by using the Ratio Test in which (and by using the Root Test in which ), provided this limit exists. At an endpoint of the interval of convergence, both the Ratio Test and Root Test always fail. Therefore, some other test should be used at the end points. (See Figure 2.)

Figure 2

EXAMPLE 1

Find the radius of convergence and interval of convergence for the series where .

Solution. Let . Then

for all

By the Ratio Test, the series converges for all . Therefore, the radius of convergence is and hence the interval of convergence is

EXAMPLE 2

Find the radius of convergence and interval of convergence for the series

.

Solution. Let  . Then

and hence,

Therefore, the series converges for all and diverges for  . Thus the radius of convergence is . When , the Ratio Test is inconclusive, so we must consider and , separately. When the series becomes the harmonic series, which diverges. When , the series is which converges by the Alternating Series Test. Thus the given power series converges when so the interval of convergence is .

EXAMPLE 3

Find the radius of convergence and interval of convergence for the series

.

Solution.  Let Then

and hence,

Therefore, the series converges for all and diverges for all Thus the radius of convergence is When the Ratio Test is inconclusive, but substituting in the series gives which is a divergent series with Setting gives which is convergent  by the Alternating Series Test. Therefore, the interval of convergence is  .

The function can be represented as a power series,

[2]

which is the geometric series with the first term and a common ratio . Certain classes of functions which may be obtained by manipulating the function may also be represented as a power series.

EXAMPLE 4

Represent the function as a power series and find its interval of convergence.

Solution. Replacing by in Equation [2], we have

.

This converges when that is, or . Thus the interval of convergence is .

EXAMPLE 5

Represent the function as a power series and find its interval of convergence.

Solution. In order to put this function in the form of the left side of Equation [2], we first factor from the denominator.

Then        .

This series converges when that is, . Thus the interval of convergence is .

EXAMPLE 6

Represent the function as a power series and find its interval of convergence.

Solution. By Example 5, Hence

Thus the interval of convergence of the series is .

The functions obtained by differentiating or integrating the functions with known power series, such as can be easily represented as a power series by using the technique called term-by-term differentiation and integration of the power series.  Any power series can be differentiated or integrated term-by-term within its interval of convergence, similar to polynomials.

THEOREM 2

Suppose  has radius of convergence Then the function is differentiable and integrable on the interval and

(ⅰ)

(ⅱ)

or equivalently

(ⅲ)

(ⅳ)

Proof.  It should be remarked that the radii of convergence of the power series in Equations (i) and (ii) remain the same. But the interval of convergence need not remain the same, that is, endpoints of the interval might change.

EXAMPLE 7

Find the radius of convergence of a power series representation of .

Solution. Differentiating both sides of the equation:

we get

.

Hence, by multiplying both sides by , we have

.

Thus the radius of convergence of this power series representation is also by Theorem 2.

EXAMPLE 8

Find a power series representation of the function and determine its radius of convergence.

Solution. The derivative of is . We know that

,  for .

Integrating both sides gives

Putting in this equation to determine the value of . Then, or . Thus,

Here since the radius of convergence is the same as for the original series. (Interval of convergence)

EXAMPLE 9

Find a power series representation of and its radius of convergence.

Solution.

Since , integrating the power series for gives

,  for .

Putting , . Therefore, for

.

The radius of convergence of this series for is 1.

CAS EXAMPLE 10

Determine the radius of convergence of the following series:

Solution.

var('n')

u(n)=1/(n*2^n)

rho=limit(abs(u(n+1)/u(n)), n=+oo)

R=1/rho; R

We may employ CAS to test convergence at the end points.

9.4 EXERCISES (Power Series)

1-10. Determine the radius of convergence and interval of convergence of the following series

1.

Solution. as .

Using the Ratio Test, the given series is absolutely convergent and therefore convergent when , and divergent when .

If , then the series becomes , which is divergent.

If , then the series becomes , which converges by the Alternating Series Test.

Thus, the given power series converges for . So, and .

2.

Solution.  as

Using the Ratio Test, the given series is absolutely convergent and therefore convergent when , and divergent when .

If , then the series becomes , which converges by the Integral Test.

If , then the series becomes , which converges by the Alternating Series Test.

Thus, the given power series converges for . Hence, and .

CAS 3.

Solution.

var('n')

u(n)=1/factorial(2*n)

rho=limit(abs(u(n+1)/u(n)), n=+oo)

rho

4.

Solution. as   .

Using the Ratio Test, the given series is absolutely convergent and therefore convergent when , and divergent when .

If , then the series becomes , which is divergent.

If , then the series becomes , which is not convergent.

Thus, the given power series converges for So, and .

5.

Solution. as .

Using the Ratio Test, the given series is absolutely convergent and therefore convergent when , and divergent when .

If , then the series becomes , which converges by the Alternating Series Test.

If , then the series becomes the negative series, which is divergent.

Thus, the given power series converges for . So, and .

6.

Solution. as .

Using the Ratio Test, the given series is absolutely convergent and therefore convergent when , and divergent when .

If , then the series becomes . Since , by the Comparison test, is convergent.

If , then the series becomes , which converges by the Alternating Series Test.

Thus, the given power series converges for . So, and .

7.

Solution. By the Root test, for all .

Thus, the radius of convergence is and the interval of convergence is .

CAS 8.

Solution.

var('n')

u(n)=1/(n^2*2^n)

rho=limit(abs(u(n+1)/u(n)), n=+oo)

R=1/rho; R

9.

Solution. as .

Using the Ratio Test, the given series is absolutely convergent and therefore convergent when , and divergent when . If , then the series becomes . Since , by the Comparison test, is divergent.

If , then the series becomes .

Since , .

Thus, the given power series converges for  . So, and .

10.

Solution. as .

Then, the given power series converges for . So, and .

11-13. Determine the interval of convergence of a power series representation for the function .

11.

Solution.

Since this is a geometric series, it converges when , that is . Therefore, the interval of convergence is .

12.

Solution.

Since this is a geometric series, it converges when . Therefore, the interval of convergence is .

13.

Solution. .

Since this is a geometric series, it converges when . Therefore, the interval of convergence is .

14. Express the function as the sum of a power series by first using partial fractions. Find the interval of convergence:

Solution.

Since this is a geometric series, it converges when and , respectively. Therefore, the interval of convergence is .

15-16. Find a power series representation for the function and determine the radius of convergence.

15.

Solution. The derivative of is .

We have , for .

Thus,

We put in this equation to determine the value of . That is, or . Thus,

Here since the radius of convergence is the same as for the original series.

16.

Solution. The derivative of is .

Now, we have

, for .

Thus,

We put in this equation to determine the value of . Then, or . Thus,

Here since the radius of convergence is the same as for the original series.

17.(a)Find a power series representation for . What is the radius of convergence?

(b)Use part (a) to find a power series for .

Solution.

(a) By Example 8 in Section 9.2,

.

(b) .

The power series for is

.

By part (a), .

Thus,

18-19. Evaluate the indefinite integral as a power series and find the radius of convergence.

18.

Solution.    .

Thus, the power series is

.

The above power series converges when . Therefore, the interval of convergence is .

19.

Solution. Since

,

.

Hence,

.

Therefore, .

9.5 Taylor, Maclaurin, and Binomial Series

Finding power series representations of functions, that is, the approximation of functions by polynomials is very useful in science and has many applications. In the preceding section we looked at a particular class of functions whose power series can be easily found by simple manipulations from a geometric series. This approach will not always work with other types of functions. In this section, we work with the class of functions that have a power series representations and determine methods to find such representations.

THEOREM 1

If has a power series representation given by

then the coefficients are given by

.

Proof.

We start by supposing that is any function that can be represented by a power series:

[1]

Let us try to determine what the coefficients must be in terms of . To begin, notice that if we put in Equation [1], then all terms after the first one are 0 and we get                                     .

We can differentiate the series in Equation 1 term by term:

[2]

and substitution of in Equation [2] gives

.

Now we differentiate both sides of Equation [2] and obtain

[3]           .

Again we put in Equation 3. The result is

.

Let us apply the procedure one more time. Differentiation of series in Equation [3] gives

[4]

and substitution of in Equation [4] gives

By now you can see the pattern. If we continue to differentiate and substitute , we obtain

Solving this equation for the th coefficient , we get

.

Theorem 1 tells us that if can be represented as a power series about , then it must be of the form

[5]  .

Note that the sum of the power series is equal to . The series on the right side is called the Taylor Series (after the English mathematician Brook Taylor, 1685–1731) of the function at (or about ).

The Taylor series of at which is given by

[6]

is called the Maclaurin Series of . (after the Scottish mathematician Colin Maclaurin, 1698–1746.)

If can be represented as a power series about , then the sum of its Taylor series is equal to In this case the value of the can be determined around based on the knowledge of and its derivatives at .

The converse of the above Theorem 1 is false: there exists a function for  which the sum of its Taylor series is not equal to the function. Consider the function

for all Therefore, the sum of its Taylor series at is

.

Hence the function is not equal to its Taylor series. Therefore, the question arises: Under what circumstances is a function equal to the sum of its Taylor series? The following theorem and corollary answer this question.

THEOREM 2   Taylor’s Theorem with Remainder

Let be times continuously differentiable on the interval . Then for all in this interval,

[7]

where for some point lying between and

The polynomial is called the th degree Taylor polynomial of at and the remainder of the Taylor series of about As an immediate consequence of the above theorem, we obtain the following important result.

THEOREM 3

Let be infinitely differentiable (smooth) on the interval . Then is equal to the sum of its Taylor series on the interval   if and only if  ,  for , where is given in Theorem 2.

Proof.

We prove necessity only. The partial sums of Taylor series are

In general, is the sum of its Taylor series if

.

Let , then . Thus

.

Therefore,   if .

Let us consider the series . The convergence of the series gives the following fact

[8]                      for every real number

which is useful in proving that a function is equal to the sum of its Taylor series.

EXAMPLE 1

Prove that sin is equal to its Maclaurin series.

Solution.  Let . Then, taking derivatives for and finding the each value at ,

In general, since ,

By Theorem 2, we have

where for some point lying between and

Thus for the remainder we have

for all values of By [8], if is any positive number, then for all values of such that

.

Consequently, ,  for any finite interval . Thus, by Corollary 3, sin is equal to its Maclaurin series for all , that is,

[9]

for all .

The graphs of sin together with its Taylor polynomials,

are shown in Figure 1. Observe that becomes a better approximation to sin as increases.

Figure 1

If is continuous on , then there exists a number in such that

.

EXAMPLE 2

Find the Maclaurin series for and prove that it is equal to for all .

Solution.  Let . Since for all , the Maclaurin series is

.

By Theorem 2, we have

where for some point lying between and   For each given , therefore,  we have

as

which implies that for any . Thus is equal to the sum of its Maclaurin series for all by Theorem 3, Hence we have

for all .

In particular, when , we obtain another expression for as a sum of an infinite series.

.

We can easily find the Taylor series expansion of about an arbitrary point since

for all .

The radius of convergence of the Maclaurin series of given by may also be found by applying the Ratio Test. Let . For any real number , we have

as .

Hence, the series converges for all by the Ratio Test with its radius of convergence .

EXAMPLE 3

Find the Maclaurin series for .

Solution.

Finding and then evaluating the derivatives of at , we get

The Maclaurin series of is, therefore,

.

As in Example 1, it can be proved that is equal to the sum of its Maclaurin series for all Thus,

for all .

EXAMPLE 4

Find the Maclaurin series for the function .

Solution.  Multiplying the Maclaurin series for sin by , we find

.

EXAMPLE 5

Find the Taylor expansion about for the function .

Solution.  The evaluation of the derivatives of at gives

Therefore the Taylor series at is

.

Some important Maclaurin series and their intervals of convergence are listed in the following table for future reference.

,

,

,

,

,

Taylor series are also useful for integrating functions that cannot be integrated explicitly using known integration techniques.

EXAMPLE 6

Evaluate as an infinite series.

Solution.  The Maclaurin series for is obtained simply by replacing with in the Maclaurin series for . Thus

for all values of . Integrate both sides term-by-term we obtained

.

This series converges for all because the original series for converges for all .

Limits can also be found using Taylor series.

EXAMPLE 7

Evaluate .

Solution.  Using the Maclaurin series for , we have

.

Power series can be added, subtracted, multiplied or divided like polynomials. However only the first few terms are calculated because the initial terms are more important and the calculations for the later terms become cumbersome.

EXAMPLE 8

Find the first three non-zero terms of the Maclaurin series for .

Solution.  Multiplying the Maclaurin series for and , then collecting like terms, we have

.

The Binomial Series

If and are any real numbers and is a positive integer, then by the Binomial Theorem, we have

This can be written in the abbreviated form as

where

and

These numbers are called the binomial coefficients.

For and , we get

[10]                                .

We extend this formula to the case where is no longer a positive integer. To attain this, we find the Maclaurin series for the function which is found to be

[11]           .

This series is called the binomial series. The following theorem gives the conditions under which the binomial expansion holds, that is, is equal to the binomial series [11], for exponent not necessarily a positive integer.

THEOREM 4   The Binomial Series Expansion

If is any real number and , then

,

where

and .

It should be noted that if is a positive integer and , then the expression for contains a factor, , so for . In this case the series terminates and reduces to the ordinary binomial expansion.

EXAMPLE 9

Using the method of binomial series expansion, expand  as a power series.

Solution.  We have and the binomial coefficient is found to be

.

By Theorem 4, for we have

.

EXAMPLE 10

Find the Maclaurin series for the function and find its radius of convergence.

Solution.  We begin by rewriting to get it into the form that we can utilize.

.

Now we can apply the binomial series with

.

This takes the following form,

This series converges when , that is, , so the radius of convergence is .

CAS EXAMPLE 11

Obtain the Taylor series for about , where , .

Solution.

var('x')

f=sin(x^2)

g=f.taylor(x, pi/4, 4)

p1=plot(f, (x, -1, 2), rgbcolor=(0, 0, 1))

p2=plot(g, (x, -1, 2), rgbcolor=(1, 0, 0))

show(p1+p2)

Figure 2

print g

Answer : 1/98304*(pi-4*x)^4 * (pi^4*sin(1/16*pi^2) - 48*pi^2 * cos(1/16*pi^2)- 192 * sin(1/16*pi^2)) + 1/3072 * (pi^3 * cos(1/16 * pi^2) + 24 * pi * sin(1/16 * pi^2)) * (pi - 4*x)^3 - 1/128*(pi^2 * sin (1/16*pi^2) - 8*cos(1/16*pi^2)) * (pi- 4*x)^2 - 1/8*(pi-4*x)*pi*cos (1/16*pi^2) + sin(1/16*pi^2)

Telescoping Series

Recall that is called a telescoping series if a sequence   , .

We note that , and .

Thus we see that converges if and only if exists, in  which case we have

.

EXAMPLE 12

Show that the series is a telescoping series and find its sum.

Solution.

var('n')

a(n)=n/(n^4+n^2+1)

an=n/(n^4+n^2+1)

s=a(n).partial_fraction()

show(s)

b(n)=1/(2*(n^2-n+1))

expand(b(n+1))

Thus if then . This implies . In particular it is a telescoping series.

var('k')

sum(a(n), n, 0, +oo)

Approximation using Taylor’s Series

Recall that if has a Taylor’s series expansion in an interval about then we have

where is the error term for some between and in and

is the degree Taylor polynomial.

Since in in the Taylor’s theorem is not known, we cannot find the error term exactly for a given in an interval . However, we can obtain a bound on the error term as

Suppose . Then we have

.

CAS EXAMPLE 13

Approximate using the first 10 Taylor polynomials.

Solution.

f(x)=sqrt(1+x)

for i in range(1, 10):

T(x)=taylor(f(x), x, 0, i)

T(x=0.1)

1.05000000000000

1.04875000000000

1.04881250000000

1.04880859375000

1.04880886718750

1.04880884667969

1.04880884829102

1.04880884816010

CAS EXAMPLE 14

Approximate using the first 10 Taylor polynomials. Note that radians.

Solution.

f(x)=sin(x)

for i in range(1, 10):

T(x)=taylor(f(x), x, 0, i)

T(x=pi/10).n()

0.314159265358979

0.314159265358979

0.308991552578929

0.308991552578929

0.309017054219328

0.309017054219328

0.309016994292875

0.309016994292875

CAS EXAMPLE 15

Find the approximate value of using the first 10 Taylor polynomials.

Solution.

f(x)=cos(x)

for i in range(1, 10):

T(x)=taylor(f(x), x, 0, i)

T(x=pi/10).n()

CAS EXAMPLE 16

Find the approximate value of .

Solution.

sin(0.2).n()

CAS EXAMPLE 17

Find the approximate value of  .

Solution.

ln(2.1).n()

6.5 EXERCISES (Taylor, Maclaurin, and Binomial Series)

1-4. Determine the radius of convergence of the Maclaurin series expansion for , where

1.

Solution.

, ,

, ,

, ,

, ,

, ,

, ,

.

Therefore, the function’s Maclaurin series is

Here , then as .

Hence, by the Ratio Test, the series converges for all , and the radius of convergence is .

2.

Solution. , for all . Therefore, Maclaurin series is

.

Let , then as .

Hence by the Ratio Test, the series converges for all , and the radius of convergence is .

3.

Solution.

, ,

, ,

, ,

, ,

, ,

.

Therefore, its Maclaurin series is

.

Here , then as .

Hence by the Ratio Test, the series converge when .

4.

Solution.

,

,

,

,

, .

Therefore, its Maclaurin series is

.

Let , then as .

Hence by the Ratio Test, the series converges for all , and the radius of convergence is .

5-8. Obtain the Taylor series for about .

5. ,

Solution.

.

6. ,

Solution.

CAS 7. ,

Solution.

var('x')

f(x)=1/(x^3)

g(x)=f.taylor(x, 1, 4)

p1=plot(f, (x, 0.5, 2))

p2=plot(g, (x, 0.5, 2), color="red", linestyle='--')

show(p1+p2)

print g

Answer : x |--> 15*(x - 1)^4 - 10*(x - 1)^3 + 6*(x - 1)^2 - 3*x + 4

CAS 8. ,

Solution.

var('x')

f(x)=log(sin(x)/x)

g(x)=f.taylor(x, 2, 3)

p1=plot(f, (x, 0, 3))

p2=plot(g, (x, 0, 3), color="red", linestyle='--')

show(p1+p2)

print g

Answer : x |--> -1/24*(x - 2)^3*(sin(2)^3 - 8*sin(2)^2*cos(2) - 8*cos(2)^3)/sin(2)^3 - 1/8*(x - 2)^2*(3*sin(2)^2 + 4*cos(2)^2)/sin(2)^2 - 1/2*(x - 2)*(sin(2) - 2*cos(2))/sin(2) - log(2) + log(sin(2)).

9-11. Find the Maclaurin series for the given function.

9. (Use .)

Solution.

,

,

,

,

,

,

Thus .

CAS 10.

Solution.

var('x')

f=(x^2)*(arctan(x))

g=f.taylor(x, 0, 4)

p1=plot(f, (x,  0.5, 6), rgbcolor=(0, 0, 1))

p2=plot(g, (x, 0.5, 6), rgbcolor=(0, 1, 0))

show(p1+p2)

print g

CAS 11.

Solution.

var('x')

f=exp(-(x^2))

g=f.taylor(x, 0, 4)

p1=plot(f, (x, 0.5, 6), rgbcolor=(0, 0, 1))

p2=plot(g, (x, 0.5, 6), rgbcolor=(0, 1, 0))

show(p1+p2)

print g

Answer : 1/2*x^4 - x^2 + 1

12-13. Evaluate the indefinite integral as an infinite series

12.

Solution.

Integrate both sides term by term:

CAS 13.

Solution.

var('t')

f=(1+(t^2))^(1/2)

g=f.taylor(t, 0, 4)

print g

Answer : -1/8*t^4 + 1/2*t^2 + 1

Thus .

Integrate both sides term by term:

14-15. Evaluate the limit using a series:

CAS 14.

Solution.

var('x')

f=log(1+x)

g=f.taylor(x, 0, 4)

h=1/(sin(x))

i=h.taylor(x, 0, 4)

j=(g^2)/(x*i-1)

p1=plot(j, (x, 0.5, 6), rgbcolor=(0, 1, 0))

show(p1)

print j

Answer: 5/2*(3*x^4 - 4*x^3 + 6*x^2 - 12*x)^2/((7*x^3 + 60*x + 360/x)*x - 360)

Thus .

15.

Solution.

,

.

16. Deduce from the Maclaurin series for that .

Solution. Since , .

Put , then .

17-20. Obtain the binomial series and radius of convergence of the function.

17.

Solution.

this series converges when , so the radius of convergence is .

18.

Solution.

this series converges when . so the radius of convergence is .

19.

Solution.

this series converges when , so the radius of convergence is .

20.

Solution.

this series converges when , so the radius of convergence is .

21. Evaluate using the binomial series where .

Solution. Since , .

Thus ,

.

Calculus