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I need to evaluate $$\sum_{n=1}^{\infty}\frac{n}{(n+2)!}$$Answer in book and WolframAlpha both say that is equal $3-e$. Thus, I have mistake and got:

$$ \begin{align} \sum_{n=1}^{\infty}\frac{n}{(n+2)!} &=0+\sum_{n=1}^{\infty}\frac{n}{(n+2)!}\\ &=\sum_{n=0}^{\infty}\frac{n}{(n+2)!}\\ &= \sum_{n=0}^{\infty}\frac{(n+2)-2}{(n+2)!}\\ &=-2\sum_{n=0}^{\infty}\frac{1}{(n+1)!}\\ &= -2(e-1)=2-2e \end{align} $$

Please help to find mistake.

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    $\begingroup$ How did you get from $\frac{(n+2)-2}{(n+2)!}$ to $-2\cdot \frac{1}{(n+1)!}$? $\endgroup$ – Daniel Fischer Apr 3 '15 at 17:35
  • $\begingroup$ I think it should be $3 - e$, just some manual labour will yield it. $\endgroup$ – user98186 Dec 29 '15 at 19:43
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$$\sum_{n=1}^\infty\frac{n+2-2}{(n+2)!}=\sum_{n=1}^\infty\frac1{(n+1)!}-2\sum_{n=1}^\infty\frac1{(n+2)!}$$

$$=\sum_{r=2}^\infty\frac1{r!}-2\sum_{u=3}^\infty\frac1{u!}$$

Now $e^x=\sum_{r=0}^\infty\dfrac{x^r}{r!}\implies e=\sum_{r=0}^\infty\dfrac1{r!}$

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  • $\begingroup$ Thanks! As both sums convergent, I can evaluate them separate? $\endgroup$ – Evgeny Egorov Apr 3 '15 at 17:42
  • $\begingroup$ @EvgenyEgorov, Definitely $\endgroup$ – lab bhattacharjee Apr 3 '15 at 17:43
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$\sum\limits_{n=1}^{\infty}\frac{n}{(n+2)!}=0+\sum\limits_{n=1}^{\infty}\frac{n}{(n+2)!}= \sum\limits_{n=0}^{\infty}\frac{n}{(n+2)!}= \sum\limits_{n=0}^{\infty}\frac{(n+2)-2}{(n+2)!}=\sum\limits_{n=0}^{\infty}\frac{1}{(n+1)!}-2\sum\limits_{n=0}^{\infty}\frac{1}{(n+2)!}=e-1-2(e-1-\frac{1}{1!})=3-e$ As expected

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