Yes, it converges

The $n$ th term of the given series is $a_{n} = \dfrac{\dfrac{n(n + 1)}{2}}{n!} = \dfrac{n + 1}{2(n - 1)!}$ .

Since $\displaystyle\lim_{n \to \infty} \dfrac{a_{n+1}}{a_{n}} = 0$ , by the ratio test this series does converge. We then have that

$\displaystyle\sum_{n=1}^{\infty} a_{n} = \sum_{n=1}^{\infty} \dfrac{n - 1 + 2}{2(n - 1)!} = \dfrac{1}{2}\sum_{n=1}^{\infty} \dfrac{n - 1}{(n - 1)!} + \sum_{n=1}^{\infty} \dfrac{1}{(n - 1)!} = \dfrac{1}{2}\sum_{n=1}^{\infty} \dfrac{n}{n!} + \sum_{n=0}^{\infty} \dfrac{1}{n!} =$

$\displaystyle\dfrac{1}{2}\sum_{n=1}^{\infty} \dfrac{1}{(n - 1)!} + e = \dfrac{1}{2}\sum_{n=0}^{\infty}\dfrac{1}{n!} + e = \dfrac{e}{2} + e = \boxed{\dfrac{3e}{2}}$ .