Summation of a series help: $\sum \frac{n-1}{n!}$ Can someone teach me how to solve the following series. I have no idea how to deal with factorial.
$$\sum_2^\infty \frac{n-1}{n!}$$
Thanks
 A: You can compute this using a few tricks:  First, since the series converges by the Ratio Test, you can split it up as the difference of two sums (each of these sums also will converge by the Ratio Test).  Then, you can re-index one of the sums to make all of the terms in the second sum line up nicely with the terms in the first, so that they all cancel (this is the idea behind telescoping sums, but is able to be written a bit more explicitly since all series involved converge), and using all of these yields a very nice result.
\begin{align*}
\sum_{n=2}^\infty \frac{n-1}{n!} &= \sum_{n=2}^\infty \frac{n}{n!} - \sum_{n=2}^\infty \frac{1}{n!}\\
&= \sum_{n=2}^\infty \frac{1}{(n-1)!} - \sum_{n=2}^\infty \frac{1}{n!}\\
&= \sum_{n=1}^\infty \frac{1}{n!} - \sum_{n=2}^\infty \frac{1}{n!}\\
&= \frac{1}{1!} = 1
\end{align*}
A: Hint: $$\frac{n-1}{n!} = \frac{1}{(n-1)!} - \frac{1}{n!}$$
A: $$\sum_2^\infty \frac{n-1}{n!}$$
$$=\sum_2^\infty \frac{1}{(n-1)!}-\frac{1}{n!}$$
As you plug in values of $n$, terms start cancelling out and the required summation is hence $=1$
