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How to evaluate the sum $\sum_{n=1}^{\infty}{n^2}/{(n+1)!}$

I know the answer to this question is $e - 1$ from wolfram alpha, but I do not understand how to get there. I know that $\sum_{n=0}^\infty \frac{1}{n!} = e,$ but I am not sure how to use it to get $$\sum_{n=1}^\infty \frac{n^2}{(n+1)!} = e-1.$$

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    $\begingroup$ Context seems to be given. They know the answer; they know the summation to use. They just need to write $n^2=(n+1)n-(n+1)+1$. This is something with which experience will help, and which might not be easy for a beginner. Help from someone with experience is worthwhile. $\endgroup$
    – robjohn
    Apr 5, 2022 at 22:44

3 Answers 3

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Although the diferentiation approach is more standard, here is an elementary method:

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

The only we need to know is that $\sum\limits_{x=0}^\infty\frac{1}{x!}=e$ and to notice that the last equality is due to a telescopic sum. The rest is elementary algebra.

Edit: I know this kind of questions doesn't have to be answered, but it was already answered when I wrote my solution.

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  • $\begingroup$ Nice elementary solution. $\endgroup$ Apr 7, 2022 at 1:58
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It is enough to express $n^2$ in a specific form:

\begin{align*} \sum_{n=1}^{\infty}\frac{n^2}{(n+1)!} &=\sum_{n=1}^{\infty}\frac{(n+1)n-(n+1)+1}{(n+1)!}\\[10pt] &=\sum_{n=1}^{\infty}\frac{(n+1)n}{(n+1)!}-\sum_{n=1}^{\infty}\frac{n+1}{(n+1)!}+\sum_{n=1}^{\infty}\frac{1}{(n+1)!}\\[10pt] &=\sum_{n=1}^{\infty}\frac{1}{(n-1)!}-\sum_{n=1}^{\infty}\frac{1}{n!}+\sum_{n=1}^{\infty}\frac{1}{(n+1)!}\\[10pt] &=\mathrm e-(\mathrm e-1)+\mathrm e-2=\mathrm e-1. \end{align*}

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  • $\begingroup$ (+1) The first step is just what the OP needs. $\endgroup$
    – robjohn
    Apr 5, 2022 at 22:45
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Changing the notation a little, $e^x =\sum_{n=0}^{\infty} \dfrac{x^n}{n!} $.

Differentiating, $e^x =\sum_{n=1}^{\infty} \dfrac{nx^{n-1}}{n!} $ so $xe^x =\sum_{n=1}^{\infty} \dfrac{nx^{n}}{n!} $.

Differentiating again, $(x+1)e^x =\sum_{n=1}^{\infty} \dfrac{n^2x^{n-1}}{n!} $ so $x(x+1)e^x =\sum_{n=1}^{\infty} \dfrac{n^2x^{n}}{n!} $.

Setting $x=1$ will give you what you need.

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  • $\begingroup$ Thank you so much $\endgroup$ Apr 5, 2022 at 19:39

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