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.
 A: 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 $x=1$.
A: Using the well-known formula for $1+2+\dots\;$your sum is 
$$\sum_{n=1}^\infty \frac{n(n+1)}{2}\frac{1}{n!}=
\sum_{n=1}^\infty \frac{(n+1)}{2(n-1)!}=
\sum_{n=0}^\infty \frac{(n+2)}{2n!}=
\sum_{n=0}^\infty \left(\frac{n}{2n!} + \frac{2}{2n!}\right)$$
$$=\frac{1}{2}\sum_{n=0}^\infty \frac{n}{n!} + \sum_{n=0}^\infty \frac{2}{2n!}$$
The first term of the first sum is zero, so we omit this term, start the summation at $n=1$ and divide out $n$ in the fraction
$$=\frac{1}{2}\sum_{n=1}^\infty \frac{1}{(n-1)!} + \sum_{n=0}^\infty \frac{1}{n!}$$
$$=\frac{1}{2}\sum_{n=0}^\infty \frac{1}{n!} + \sum_{n=0}^\infty \frac{1}{n!}$$
$$=\frac{1}{2}e + e = \frac{3}{2}e$$
A: Hint:
For any polynomial $P(x)$, with $D$ denoting the differential operator $\frac{d}{dx}$,
$$P(x D) e^x =  \sum_{n=0}^\infty \frac{P(n)}{n!} x^n$$
Here $P(x) = \frac12x(x+1)$, so we need $\frac12 xD(xD+1) e^x$, which gives $\frac12xe^x(x+2)$, which on setting $x=1$ gives $\frac32e$.
