# Finding radius and interval of convergence

• Nov 30th 2011, 06:28 PM
Intrusion
Finding radius and interval of convergence
The problem: Find the radius and interval of convergence.
$$\sum_{1}^{\infty}\frac{2^n(x-2^n)}{(n+2)!}$$

The attempt:
I basically did the ratio test and it equaled 0. So does R = $\infty$ and I = $(-\infty,\infty)$?
• Nov 30th 2011, 07:20 PM
chisigma
Re: Finding radius and interval of convergence
Quote:

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

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

If there are no errors, the expression can be written as...

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

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

Kind regards

$\chi$ $\sigma$
• Dec 1st 2011, 03:38 AM
HallsofIvy
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
$\sum_{n=0}^\infty\frac{2^n(x- 2)^n}{(n+2)!}$

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