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I am trying to find the explanation and proof of the sum till inifinity.

$1 + \frac{1+2} {2!} + \frac{1+2+2^2} {3!} + \frac{1+2+2^2+2^3} {4!}$ + . . . .

I know the explanation of the more simplier series like below

$ 1+ \frac{1+2}{2!}+ \frac{1+2+3}{3!}+ \frac{1+2+3+4}{4!} ….$

The explanation is,

The general term of the series can be given as
$T_n=1$ for $n=1$ and for $n\geq 2$ we have, $T_n=\dfrac{\sum_{i=1}^ni}{i!}$ $=\dfrac{i(i+1)}{2i!}$
$=\dfrac{i+1}{2(i-1)!}$ $=\dfrac{1}{2(i-2)!}+\dfrac{1}{(i-1)!}.$

Sum of the series= $\sum_{i=1}^\infty T_i=1+\sum_{i=2}^\infty > \left(\dfrac{1}{2(i-2)!}+\dfrac{1}{(i-1)!}\right)$ All the terms are positive, we can rearrange the terms $=1+\dfrac{1}{2}\left(1+\dfrac{1}{1!}+\ldots\right)+\left(\dfrac{1}{1!}+\dfrac{1}{2!}\ldots\right)$ $=\dfrac{e}{2}+e=\dfrac{3e}{2}$

But what will be the summation of above series ? It would be better if i get proof and explanation too

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  • $\begingroup$ Hint: compute $1 + 2 + 2^2 + \cdots + 2^n$. $\endgroup$ – Gribouillis Sep 10 '17 at 14:31
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$$\sum_{n=1}^\infty\frac1{n!}\sum_{k=0}^{n-1}2^k=\sum_{n=1}^\infty\frac1{n!}(2^n-1)=\sum_{n=1}^\infty\frac{2^n}{n!}-\sum_{n=1}^\infty\frac{1^n}{n!}=(e^2-1)-(e-1)=e^2-e$$

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