Infinite Factorial Series

$\displaystyle \sum_{n=1}^{\infty} \dfrac{n}{(n + 1)!} = \sum_{n=1}^{\infty} \dfrac{(n + 1) - 1}{(n + 1)!} = \sum_{n=1}^{\infty} \left(\dfrac{n + 1}{(n + 1)!} - \dfrac{1}{(n + 1)!} \right) = \sum_{n=1}^{\infty} \left(\dfrac{1}{n!} - \dfrac{1}{(n + 1)!}\right) = \boxed{1}$ ,

as the last sum telescopes so that only the first element of the first term remains.

$\begin{aligned}\sum_{n=0}^\infty \frac{x^n}{n!}&=e^x\\ \sum_{n=1}^\infty \frac{x^n}{n!}&=e^x-\frac{x^0}{0!}\\ \sum_{n=0}^\infty \frac{x^{n+1}}{(n+1)!}&=e^x-1\\ \sum_{n=0}^\infty \frac{x^n}{(n+1)!}&=\frac{e^x-1}{x}\\ \sum_{n=1}^\infty \frac{x^n}{(n+1)!}&=\frac{e^x-1}{x}-\frac{x^0}{(0+1)!}\\ \frac{d}{dx}\left(\sum_{n=1}^\infty \frac{x^n}{(n+1)!}\right)&=\frac{d}{dx}\left(\frac{e^x-1}{x}-1\right)\\ \sum_{n=1}^\infty \frac{nx^{n-1}}{(n+1)!}&=\frac{xe^x-(e^x-1)}{x^2}\\ \underline{\text{let }x=1:}\\ \sum_{n=1}^\infty \frac{n(1)^{n-1}}{(n+1)!}&=\frac{1e^1-(e^1-1)}{1^2}\\ \sum_{n=1}^\infty \frac{n}{(n+1)!}&=1\quad\blacksquare\end{aligned}$