TITLE: Where does this series converges $1+\frac{1+2}{2!}+\frac{1+2+3}{3!}+\cdots$
QUESTION [8 upvotes]: Consider the following series,

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

My Efforts
The series can be written in compact for as, 
$$\sum_{n=1}^{\infty}{n(n+1)\over 2n!}$$
$$={1\over 2}(\sum_{n=1}^{\infty}{n^2\over n!}+\sum_{n=1}^{\infty}{n \over n!})$$
$$={1\over 2}(\sum_{n=1}^{\infty}{n^2\over n!}+\sum_{n=1}^{\infty}{1 \over (n-1)!})$$
Now 
$$\sum_{n=1}^{\infty}{1 \over (n-1)!}=1+1+{1\over 2!}+{1\over 3!}+\cdots=e$$
First summand can be simplified as, 
$$\sum_{n=1}^{\infty}{n \over (n-1)!}$$
But where does it converge? Any hints?
Edit: Thank you for all the quick replies. I now realize it was so easy. I just had to use $n=n-1+1$.

REPLY [3 votes]: \begin{align}
\sum_{n=1}^{\infty}{n(n+1)\over 2n!}
&= \sum_{n=1}^{\infty}{n-1+2\over 2(n-1)!} \\
&= \sum_{n=2}^{\infty}\dfrac{1}{2(n-2)!}+\sum_{n=1}^{\infty}\dfrac{1}{(n-1)!} \\
&= \dfrac{1}{2}e+e \\
&= \color{blue}{\dfrac{3}{2}e}
\end{align}

REPLY [2 votes]: $\sum_{n=1}^{\infty}\frac{n}{\left(n-1\right)!}=1+\sum_{n=2}^{\infty}\frac{\left(n-1\right)+1}{\left(n-1\right)!}=1+\sum_{n=0}^{\infty}\frac{1}{n!}+\sum_{n=1}^{\infty}\frac{1}{n!}=1+e+(e-1)=2e$
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