Possible Duplicate:
Finding the sum of this alternating series with factorial denominator.
Find$\sum_{0}^{\infty}\frac{(-1)^n(n+1)}{n!}$ I have tried Taylor series, but it does not have $x^n$. Is there any trick to solve the problem?
Possible Duplicate:
Finding the sum of this alternating series with factorial denominator.
Find$\sum_{0}^{\infty}\frac{(-1)^n(n+1)}{n!}$ I have tried Taylor series, but it does not have $x^n$. Is there any trick to solve the problem?
\begin{align} S_N & = \sum_{n=0}^{N} (-1)^n \dfrac{(n+1)}{n!}\\ & = \sum_{n=0}^{N} (-1)^n \left(\dfrac{n}{n!} + \dfrac1{n!} \right)\\ & = \sum_{n=0}^{N} (-1)^n \dfrac{n}{n!} + \sum_{n=0}^{N} (-1)^n \dfrac1{n!}\\ & = \sum_{n=1}^{N} (-1)^n \dfrac1{(n-1)!} + \sum_{n=0}^{N} (-1)^n \dfrac1{n!}\\ & = \left( -\dfrac1{0!} + \dfrac1{1!} - \dfrac1{2!} + \cdots + \dfrac{(-1)^n}{(n-1)!}\right) + \sum_{n=0}^{N} (-1)^n \dfrac1{n!}\\ & = \sum_{n=0}^{N-1} (-1)^{n+1} \dfrac1{n!} + \sum_{n=0}^{N} (-1)^n \dfrac1{n!}\\ & = \dfrac{(-1)^N}{N!} \end{align} Can you now finish it off by letting $N \to \infty$?
$\begin{align*}\sum_{n\ge 0}\frac{(-1)^n(n+1)}{n!}&=\sum_{n\ge 0}\frac{(-1)^nn}{n!}+\sum_{n\ge 0}\frac{(-1)^n}{n!}\\ &=\sum_{n\ge 1}\frac{(-1)^n}{(n-1)!}+\sum_{n\ge 0}\frac{(-1)^n}{n!}\\ &=\sum_{n\ge 0}\frac{(-1)^{n+1}}{n!}+\sum_{n\ge 0}\frac{(-1)^n}{n!}\\ &=\sum_{n\ge 0}\frac{(-1)^{n+1}+(-1)^n}{n!}\;, \end{align*}$
which is what?