Finding radius and interval of convergence

**Intrusion** · November 30th 2011, 06:28 PM

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 = $(-\infty,\infty)$ ?

**chisigma** · November 30th 2011, 07:20 PM

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

$\chi$ $\sigma$

**HallsofIvy** · December 1st 2011, 03:38 AM

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.

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