Calculus-Sec-9-4-Solution

9.4   Power Series                                   by SGLee - HSKim -JHLee

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

1.

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.

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 .

4.

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.

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.

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 Sereis Test.

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

7.

By the Root test,

for all .

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

9.

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.

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.

Since this is a geometric series, it converges when , that is .

Therefore, the interval of convergence is .

12.

Since this is a geometric series, it converges when .

Therefore, the interval of convergence is .

13.

.

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 :  .

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.

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.

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 .

(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.

.

Thus, the power series is

.

The above power series converges when .

Therefore, the interval of convergence is .

19.

Since

.

Hence,

.

Therefore,

.

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