Answer
Converges
Work Step by Step
Since, we have $a_n=\Sigma_{n=2}^\infty \dfrac{1}{n !}=\Sigma_{n=2}^\infty \dfrac{1}{n (n-1)(n-2)(n-3)!}$
and $a_n \lt \Sigma_{n=2}^\infty \dfrac{1}{n (n-1)!} \lt \Sigma_{n=2}^\infty \dfrac{1}{(n-1)(n-1)}=\Sigma_{n=1}^\infty \dfrac{1}{n^2}$
we can see that $\Sigma_{n=2}^\infty \dfrac{1}{n !} \lt \Sigma_{n=1}^\infty \dfrac{1}{n^2}$
which is a p-series with $p=2 \gt 1$ and thus, convergent.
Hence, the given series also converges.
