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


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

Can you explain it, please ?.

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  • $\begingroup$ @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 $\mbox{}\displaystyle{\quad \sum_{n = 1}^{\infty}{2^{n}\left[\log\left(2\right)\right]^{n} \over n!}}$. :) $\endgroup$ – 1 0 Nov 12 '16 at 18:12
  • $\begingroup$ Originally, it was written different. $\endgroup$ – Felix Marin Nov 13 '16 at 20:04
  • $\begingroup$ @Felix Marin, No, that was correct, you can see :) math.stackexchange.com/posts/2010300/revisions $\endgroup$ – 1 0 Nov 13 '16 at 21:05
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Using the exponential series, you find

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

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

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  • 1
    $\begingroup$ @MithleshUpadhyay You are welcome. $\endgroup$ – Olivier Oloa Nov 12 '16 at 9:25

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