Answer
Converges
Work Step by Step
Let $u_n=\Sigma_{n=2}^\infty \dfrac{1}{n !}$
Re-write the given series such as: $\Sigma_{n=2}^\infty \dfrac{1}{n !}=\Sigma_{n=2}^\infty \dfrac{1}{n (n-1)(n-2)(n-3)!}$
We compare with: $ \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})$
This implies that $\Sigma_{n=2}^\infty (\dfrac{1}{n !}) \lt \Sigma_{n=1}^\infty (\dfrac{1}{n^2})$
and $\Sigma_{n=1}^\infty \dfrac{1}{n^2}$ shows a convergent $p$-series with $p=2 \gt 1$ .
Thus, the series converges by the limit comparison test.
