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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?

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Jill Clover
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$$\begin{align*}\sum_{n\ge 0}\frac{(-1)^n(n+1)}{n!}&=\sum_{n\ge 0}\frac{(-1)^nn}{n!}+\sum_{n\ge 0}\frac{(-1)^n}{n!}\\ &=\sum_{n\ge 1}\frac{(-1)^n}{(n-1)!}+\sum_{n\ge 0}\frac{(-1)^n}{n!}\\ &=\sum_{n\ge 0}\frac{(-1)^{n+1}}{n!}+\sum_{n\ge 0}\frac{(-1)^n}{n!}\\ &=\sum_{n\ge 0}\frac{(-1)^{n+1}+(-1)^n}{n!}\;, \end{align*}$$

which is what?

Brian M. Scott
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\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 $N \to \infty$?

  • Your original answer was better... – wj32 Oct 28 '12 at 06:01
  • @wj32 But that had to do with manipulating infinite series, though it was legal. And anyways I do not see much difference between the current answer and the previous one I had. –  Oct 28 '12 at 06:01
  • @Marvis From the 4th column to 5th column, how does$(n-1)! $becomes $n$.By substitute? Thanks : ) – Jill Clover Oct 28 '12 at 07:19
  • @PENGTENG Yes. It is just a change of variables $n \to n+1$. I have expanded it out to be clear. –  Oct 28 '12 at 07:22