Is there a nice way to see that the series $$\sum_{n=0}^\infty (-1)^n (n+1)/n!$$ converges to $0$? I just did the computation numerically. Thanks
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$$\eqalign{ & \sum\limits_{n = 0}^\infty {{{( - 1)}^n}} \frac{{n + 1}}{{n!}} = \cr & = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}n}}{{n!}}} + \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}}}{{n!}}} = \cr & = \sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^n}}}{{\left( {n - 1} \right)!}}} + \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}}}{{n!}}} = \cr & = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^{n + 1}}}}{{n!}}} + \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}}}{{n!}}} = - \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}}}{{n!}}} + \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}}}{{n!}}} = 0 \cr} $$ |
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First notice that $$e^{-x}=\sum_{n=0}^\infty \frac{(-1)^nx^n}{n!}.$$ If we multiply this by $x$, we get $$xe^{-x}=\sum_{n=0}^\infty \frac{(-1)^nx^{n+1}}{n!}.$$ Now if we differentiate, we get $$e^{-x}-xe^{-x}=\sum_{n=0}^\infty \frac{(-1)^n (n+1)x^{n}}{n!}.$$ We now set $x=1$ to get $$0=e^{-1}-e^{-1}=\sum_{n=0}^\infty \frac{(-1)^n (n+1)}{n!}.$$ |
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