<s>[INST] Convergence of infinite series question The sum of the series 
$1+\frac{1+2}{1!}+\frac{1+2+3}{3!}+....$ equals?
The answer is $\frac{3e}{2}$.
But I dont Know How?
I have tried following:
$1+\frac{1+2}{1!}+\frac{1+2+3}{3!}+....$=$\sum\limits_{n=1}^{\infty} \frac{n(n+1)}{2n!}
$
I know that sequence of partial sum $S_n$=$1+\frac{1+2}{1!}+\frac{1+2+3}{3!}+....+\frac{n(n+1)}{2n!}$ and $\sum\limits_{n=1}^{\infty} \frac{n(n+1)} {2n!}$ converges to same point.
Now how to find $S_n$ converges to which point?
 [/INST] Hint:
$$\frac{n(n+1)}{n!}=\frac{n+1}{(n-1)!}=\frac{(n-1)+2}{(n-1)!}=\frac{(n-1)}{(n-1)!}+2\frac{1}{(n-1)!}=\frac{1}{(n-2)!}+2\frac{1}{(n-1)!}$$
the both last ones with $n\to\infty$ gives us $e$
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