Question: Find $\sum\_{n=1}^{\infty}(-1)^n\frac{n}{(n+1)!}$

**here's first attempt**

\begin{eqnarray}
\sum\_{n=1}^{\infty}(-1)^n\frac{n}{(n+1)!}&=&\sum\_{n=1}^{\infty}(-1)^{n-1}\frac{n-1}{(n)!}\\
&=&\sum\_{n=1}^{\infty}(-1)^{n}\frac{1-n}{(n)!}\\
&=&\sum\_{n=1}^{\infty}(-1)^{n}\frac{1}{(n)!}-\sum\_{n=1}^{\infty}(-1)^{n}\frac{n}{(n)!}
\end{eqnarray}
By $$e^{-x}=\sum\_{n=0}^{\infty}(-1)^n\frac{1}{n!}x^n$$we have
$$e^{-1}-1=\sum\_{n=1}^{\infty}(-1)^{n}\frac{1}{(n)!}$$
so we get $$\sum\_{n=1}^{\infty}(-1)^n\frac{n}{(n+1)!}=2e^{-1}-1$$

What are other possible approaches?
thanks

Answer: 
$$\frac{e^x -1}{x}=\sum\_0^{\infty} \frac{x^{n}}{(n+1)!} $$
Differentiate:

$$\frac{xe^x -(e^x-1)}{x^2}=\sum\_0^{\infty} \frac{nx^{n-1}}{(n+1)!}$$

Set $x=-1$:

$$1-2e^{-1} = \sum\_1^{\infty} \frac{(-1)^{n-1} n}{(n+1)!} $$

Multiply both sides by $-1$:

$$2e^{-1}-1 =\sum\_1^{\infty} \frac{(-1)^n n}{(n+1)!} $$
Almots the same as you did, maybe a bit clearer.

We have $\frac{n}{(n+1)!} = \frac{1}{n!}\cdot\frac{n}{n+1}=\frac{1}{n!}\cdot (1-\frac{1}{n+1}) = \frac{1}{n!}-\frac{1}{(n+1)!}$.

So
$$\sum\limits\_{n=1}^{+\infty} \frac{n}{(n+1)!} = \sum\limits\_{n=1}^{+\infty} \frac{(-1)^n}{n!} - \sum\limits\_{n=1}^{+\infty} \frac{(-1)^n}{(n+1)!} = \sum\limits\_{n=0}^{+\infty} \frac{(-1)^n}{n!} - 1 + \sum\limits\_{n=0}^{+\infty} \frac{(-1)^n}{n!} -1+1$$
$$=2e^{-1} -1$$