Where does this series converges $1+\frac{1+2}{2!}+\frac{1+2+3}{3!}+\cdots$

Hint:

$$ \sum_{n=1}^{\infty}{n \over (n-1)!} = \sum_{n=1}^{\infty}{n - 1 + 1 \over (n-1)!} $$

$$ = \sum_{n=2}^{\infty}{1 \over (n-2)!} + \sum_{n=1}^{\infty}{ 1 \over (n-1)!} $$

\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}

Hint: $\sum_{k=1} \frac{n^2}{n!} x^n =e^x (x^2 + x)$.

This can easily be proven by considering the Taylor expansion of $e^x$, multiplying, and making the change of variable.