# Find $\sum_{n = 1}^{\infty}{2^{n}\left[\log\left(2\right)\right]^{n} \over n!}$

$$\mbox{Find}\displaystyle{\quad \sum_{n = 1}^{\infty}{2^{n}\left[\log\left(2\right)\right]^{n} \over n!}}$$.

Answer is $$3$$, by the ratio test the series converges. I googled it, but stuck for procedure.

Can you explain it, please ?.

• @Felix Marin, oops, that's not $\mbox{}\displaystyle{\quad \sum_{n = 1}^{\infty}{\left[2^{n}\log\left(2\right)\right]^{n} \over n!}}$, It's given as $\mbox{}\displaystyle{\quad \sum_{n = 1}^{\infty}{2^{n}\left[\log\left(2\right)\right]^{n} \over n!}}$. :) – 1 0 Nov 12 '16 at 18:12
• Originally, it was written different. – Felix Marin Nov 13 '16 at 20:04
• @Felix Marin, No, that was correct, you can see :) math.stackexchange.com/posts/2010300/revisions – 1 0 Nov 13 '16 at 21:05

## 2 Answers

Using the exponential series, you find

$$\sum_{n = 1}^{\infty}{2^{n}\left[\log\left(2\right)\right]^{n} \over n!}=e^{2\ln 2 }-1=2^2-1=3$$

One may recall that $$\sum_{n=0}^{\infty}\frac{x^{n}}{n!}=e^x,\quad x \in \mathbb{R}.$$

• @MithleshUpadhyay You are welcome. – Olivier Oloa Nov 12 '16 at 9:25