# Sum of series $\frac{n}{(n+1)!}$

I'm encountering some difficulty on a question for finding the sum of the series $$\sum_{n=0}^\infty \dfrac{n}{(n+1)!}$$

The method I use to tackle this type of problem is generally to find a similar sum of a power series and algebraically manipulate it to match that of the original. I haven't found anything similar except for the summation of $e^x$ starting from $n=-1$, and subbing in $n^{\frac{1}{n}}$. Though, I'm not sure that will even work.

• Since $\frac{n}{(n+1)!} = \frac{1}{n!} - \frac{1}{(n+1)!}$, this is a telescoping series! Nov 20, 2013 at 5:55
• See Dobinski's formula for Bell numbers : $$B_k=\frac1e\cdot\sum_{n=0}^\infty\frac{n^k}{n!}\in\mathbb{N}\quad\forall\ k\in\mathbb{N}$$ Nov 20, 2013 at 6:50
$$\sum_{n=0}^\infty \dfrac{n}{(n+1)!}= \sum_{n=1}^\infty \dfrac{n-1}{(n)!}$$
We have $$\sum_{n=0}^\infty \dfrac{n}{(n+1)!} = \sum_{n=0}^\infty \left (\frac {1} {n!} - \frac {1} {(n + 1)!} \right) = \frac {1} {0!} - \frac {1} {\infty!} = 1 - 0 = 1.$$