1. ## Finding radius and interval of convergence

The problem: Find the radius and interval of convergence.
$\displaystyle $\sum_{1}^{\infty}\frac{2^n(x-2^n)}{(n+2)!}$$

The attempt:
I basically did the ratio test and it equaled 0. So does R = $\displaystyle \infty$ and I = $\displaystyle (-\infty,\infty)$?

2. ## Re: Finding radius and interval of convergence

Originally Posted by Intrusion
The problem: Find the radius and interval of convergence.
$\displaystyle $\sum_{1}^{\infty}\frac{2^n(x-2^n)}{(n+2)!}$$

The attempt:
I basically did the ratio test and it equaled 0. So does R = $\displaystyle \infty$ and I = $\displaystyle (-\infty,\infty)$?
If there are no errors, the expression can be written as...

$\displaystyle f(x)= x\ \sum_{n=1}^{\infty} \frac{2^{n}}{(n+2)!} - \sum_{n=1}^{\infty} \frac{4^{n}}{(n+2)!}= a x + b$ (1)

... so that the x doesn't play any role in convergence...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. ## Re: Finding radius and interval of convergence

chisigma's point is that, the way you wrote it here, that is NOT a power series. I expect that you meant
$\displaystyle \sum_{n=0}^\infty\frac{2^n(x- 2)^n}{(n+2)!}$

Yes, the limit of the ratios is 0 which is less than 1 for all x. The "radius of convergence" is infinite.