Hello, I need to check my answer:

inf
Sum
n=1

(-1)^n * x^(2n+1)
_________________
(2n+1)!

I preform the ratio test, use bn+1/bn...replace alln with n+1...times by reciprocal, etc...the typical method and get:

1/2|x^2/(n+1)(2n+3)|

Taking the limit, the answer = 0....thus the radius is then INFINTE.

So the radius is Infinity. If we were asked, then the int of convergence is (-inf, inf).

2. Originally Posted by orendacl
Hello, I need to check my answer:

inf
Sum
n=1

(-1)^n * x^(2n+1)
_________________
(2n+1)!

I preform the ratio test, use bn+1/bn...replace alln with n+1...times by reciprocal, etc...the typical method and get:

1/2|x^2/(n+1)(2n+3)|

Taking the limit, the answer = 0....thus the radius is then INFINTE.

So the radius is Infinity. If we were asked, then the int of convergence is (-inf, inf).

I'm not sure what you are asking. Are you asking if the interval is infinite or how to write your answer? The interval is infinite.

3. Yes, What is the radius of convergence. I do not need to state the interval but I thought as listed in my reasoning (that the interval was -inf to inf).
Can you show how you got a finite answer?

As said, I used ration test. Then replace all n with n+1. Set up bn+1/bn
If I work this out then I find the infinite solution for the radius. I then thought that the interval was not finite based upon the radius. Is the radius also not finite?

4. Originally Posted by orendacl
Yes, What is the radius of convergence. I do not need to state the interval but I thought as listed in my reasoning (that the interval was -inf to inf).
Can you show how you got a finite answer?

As said, I used ration test. Then replace all n with n+1. Set up bn+1/bn
If I work this out then I find the infinite solution for the radius. I then thought that the interval was not finite based upon the radius. Is the radius also not finite?
Your correct the interval is infinity. Some functions like the Bessel function have this characteristic

5. In fact, that power series is the series for sin(x). Since sin(x) is analytic everywhere, its power series converges everywhere.