10.4 Power Series

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

1. .

Sol) 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 .

2. .

Sol) 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 Sereis Test.

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

3. .

http://matrix.skku.ac.kr/cal-lab/cal-10-4-3.html

Sol)

var('n') u(n)=1/factorial(2*n) rho=limit(abs(u(n+1)/u(n)), n=+oo) rho |

0

4. .

Sol) 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 convergent.

If , then the series becomes , which is convergent.

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

5. .

Sol) as .

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

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

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

6. .

Sol) as .

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. .

Sol) By the Root test,

for all .

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

8. .

http://matrix.skku.ac.kr/cal-lab/cal-10-4-8.html

Sol)

var('n') u(n)=1/(n^2*2^n) rho=limit(abs(u(n+1)/u(n)), n=+oo) R=1/rho; R |

2

9. .

Sol) as .

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. .

Sol) 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. .

Sol)

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

12. .

Sol)

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

13. .

Sol) .

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 : .

Sol)

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. .

Sol) 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. .

Sol) 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?

Sol) The derivative of is . We know that

, 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.

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

Sol) By part (a),

.

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

Sol) .

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. .

Sol) .

Thus, the power series is

.

It converges when . Therefore, the interval of convergence is .

19. .

Sol) Since

.

Hence,

.

Therefore,

.