TITLE: Find the summation $\frac{1}{1!}+\frac{1+2}{2!}+\frac{1+2+3}{3!}+ \cdots$
QUESTION [11 upvotes]: What is the value of the following sum?

$$\frac{1}{1!}+\frac{1+2}{2!}+\frac{1+2+3}{3!}+ \cdots$$

The possible answers are:

A. $e$
B. $\frac{e}{2}$
C. $\frac{3e}{2}$
D.  $1 + \frac{e}{2}$

I tried to expand the options using the series representation of $e$ and putting in $x=1$, but I couldn't get back the original series. Any ideas?

REPLY [8 votes]: $$\frac{k(k+1)}{2k!}=\frac{k(k-1)+2k}{2k!}=\frac1{2(k-2)!}+\frac1{(k-1)!}$$
Hence the sum is
$$\frac e2+e.$$

Note that the first summation in the RHS must be started at $k=2$ (or use $1/(-1)!:=0$).