## Calculus (3rd Edition)

Given $$\sum_{n=1}^{\infty} \frac{1}{(n+1) !}$$ Compare with $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{2}}$, which is a convergent series ( $p-$series with $p=2$) and for $n\geq 1$ \begin{align*} (n+1) ! &\geq n^{2}\\ \frac{1}{(n+1) !} &\leq \frac{1}{n^{2}} \end{align*} Then $\displaystyle\sum_{n=1}^{\infty} \frac{1}{(n+1) !}$ also converges.