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.

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
$$=\frac{1}{2}\sum_{n=1}^\infty \frac{1}{(n-1)!} + \sum_{n=0}^\infty \frac{1}{n!}$$
$$=\frac{1}{2}\sum_{n=0}^\infty \frac{1}{n!} + \sum_{n=0}^\infty \frac{1}{n!}$$
$$=\frac{1}{2}e + e = \frac{3}{2}e$$
