Finding the sum of a specific alternating series The series $$\sum_{n=1}^\infty \dfrac{(-1)^{n +1} (n+1)}{n!}$$ can be easily proven by using Leibniz's test to be convergent. But I am finding a problem in finding it's sum.
 Some help is much required. Thank you
 A: $$(-1)^{n+1}\cdot\dfrac{n+1}{n!}=\dfrac{(-1)^{n-1}}{(n-1)!}-\dfrac{(-1)^n}{n!}=f(n-1)-f(n)$$
where $f(m)=\dfrac{(-1)^m}{m!}$
See https://en.m.wikipedia.org/wiki/Telescoping_series
A: Note that, since all of the relevant sums converge absolutely
$$\sum\limits_{i=1}^\infty \frac{(-1)^{n+1}(n+1)}{n!} = \sum\limits_{i=1}^\infty\left(\frac{(-1)^{n+1}}{(n-1)!} + \ \frac{(-1)^{n+1}}{n!}\right).$$
Notice that (again, since everything in sight converges absolutely) the second half of the $n$th term cancels with the first half of the $(n+1)$st term, so our sum telescopes to 
$$\frac{(-1)^2}{0!} = 1.$$
A: $
\sum_{n=1}^{\infty}\frac{(-1)^{n+1}(n+1)}{n!}\\
=\sum_{n=0}^{\infty}\frac{(-1)^{n}(n+2)}{(n+1)!}\\
=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!}+\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(n+1)!}\\
=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!}+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n!}\\
=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!}-\sum_{n=1}^{\infty}\frac{(-1)^n}{n!}\\
=1
$
A: Starting with 
$$e^{-x} = \sum_{n=0}^{\infty} (-1)^n \frac{1}{n!} x^n$$
Multiply by $x$
$$x e^{-x} = \sum_{n=0}^{\infty} (-1)^n \frac{1}{n!} x^{n+1}$$
Differentiate with respect to $x$
$$e^{-x}-x e^{-x} = \sum_{n=0}^{\infty} (-1)^n \frac{n+1}{n!} x^n = 1 - \sum_{n=1}^{\infty} (-1)^{n+1} \frac{n+1}{n!} x^n$$
Now set $x=1$, with result
$$0 = 1 - \sum_{n=1}^{\infty} (-1)^{n+1} \frac{n+1}{n!} x^n$$
so the sum is $1$.
