interval of convergence for power series

[sambellamy]

I am having trouble with a problem. The question asks the interval of convergence for ∑^∞_n=1 (2n)! (xⁿ) / 2ⁿ. I used the ratio test, with

a_n+1= (2n+2)(2n+1)(2n)!(xⁿ)(x) / (2ⁿ)(2) and I got some things to cancel, and came up with:

lim_n->∞ | (2n+2)(2n+1)(x) / 2 |

= lim_n->∞ |x| (2n+2)(2n+1)(1/2)

and I am not sure where to go from here. I know the absolute value of x must be < 1 for convergence, but I see here that if n were to grow very large this statement would grow larger than 1. I also know that a power series must be convergent at least at a for (x-a), so this would be zero for this series, but I don't see how that will happen as x is larger than 1. Please help!

 

[HallsofIvy]

Perhaps it was a mistype but "x is greater than 1" has nothing to do with this. What is true is that, as n goes to infinity, the whole thing goes to infinity for all x except x= 0. The interval of convergence is simply [0, 0]= {0}. The "radius of convergence" is 0 and the series converges only for x= 0.

 

[sambellamy]

Ah, writing it out like this and reasoning that we have to figure out which x values would make this do what makes a lot more sense. Thank you for the insight!
There was some part of the ratio test where the radius of convergence is that which makes either the absolute value of x, or the absolute value of the whole thing less than 1. Am I getting two tests confused?

 

[HallsofIvy]

The ratio test says that the series \(\displaystyle \sum a_n\) converges absolutely if \(\displaystyle \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|< 1\). In the case of a power series, say \(\displaystyle \sum a_nx^n\) that is \(\displaystyle \left|\frac{a_{n+1}x^{n+1}}{a_nx^n}\right|= \left|\frac{a_{n+1}}{a_n}\right| |x|< 1\). There is a |x| in that but it is the "whole thing" that must be less than 1, not just the |x|.

 

[sambellamy]

Thanks for the clarification!