The value of _k=1^ (-1)^k+1(k(k+1))/(k!) is equal to:

Question

The value of $\displaystyle\sum_{k=1}^{\infty} (-1)^{k+1}\frac{k(k+1)}{k!}$ is equal to:

Accepted Answer

For $k\ge2$, $\dfrac{k(k+1)}{k!} = \dfrac{k+1}{(k-1)!} = \dfrac{(k-1)+2}{(k-1)!} = \dfrac{1}{(k-2)!}+\dfrac{2}{(k-1)!}$.

$$S = \sum_{k=2}^{\infty}\frac{(-1)^{k+1}}{(k-2)!} + \sum_{k=1}^{\infty}\frac{2(-1)^{k+1}}{(k-1)!}.$$

Using $e^{-1} = \displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}$: substituting $n=k-2$ in the first sum gives $-\tfrac1e$, and $m=k-1$ in the second gives $\tfrac2e$.

$$S = -\frac1e + \frac2e = \frac1e.$$

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