<s>[INST] Calculate the sum of the infinite series $ 1 + \frac{1+2}{2!} + \frac{1+2+3}{3!} .... $ Calculate the sum of the infinite series $ 1 + \frac{1+2}{2!} + \frac{1+2+3}{3!} .... $
My attempt : I recognised that this series can be decomposed into the taylor expansion of $ e $ around 0. 
So I thought of writing the series as 
$ 1 + \frac{1}{2!} + \frac{1}{3!} ...$ + $ 2[ \frac{1}{2!} + \frac{1}{3!}...]$ $ +$ $ 3[ \frac{1}{3!} + \frac{1}{4!} ...] + ...$
However I got stuck here and couldn't proceed further. 
Any hints on how to proceed further , or a better method to solve the question would be appreciated.
 [/INST] The general term is $$a_n=\frac{1}{n!}\sum_{i=1}^n i=\frac 12\frac{n (n+1)}{ n!}$$ So, now consider $$S=\frac 12\sum_{n=1}^\infty \frac{n (n+1)}{ n!}x^n=\frac 12\sum_{n=1}^\infty \frac{n (n-1)+2n}{ n!}x^n=\frac 12\sum_{n=1}^\infty \frac{n (n-1)}{ n!}x^n+\sum_{n=1}^\infty \frac{n}{ n!}x^n$$ $$S=\frac {x^2}2\sum_{n=1}^\infty \frac{n (n-1)}{ n!}x^{n-2}+x\sum_{n=1}^\infty \frac{n}{ n!}x^{n-1}$$ where you should recognize  derivatives of  $e^x$. At the end, set $x=1$.
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