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The summation in question is $$\sum_{n=0}^\infty \frac{n(n+1)(n+2)}{n! + (n+1)! + (n+2)!}$$

The sum can be simplified further into $$\sum_{n=0}^\infty \frac{n(n+1)^2}{(n+2)!}$$ With Taylor expansion allowed, I don't think it's hard to derive it from expansion of $e^x$.

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    $\begingroup$ With or Without ? $\endgroup$ Feb 6, 2018 at 4:38

2 Answers 2

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As $n(n+1)^2$ is $O(n^3),$

write $n(n+1)^2=(n+2)(n+1)n+ A(n+2)(n+1)+B(n+2)+C$

so that $$\dfrac{n(n+1)^2}{(n+2)!}=\dfrac1{(n-1)!}+\dfrac A{n!}+\dfrac B{(n+1)!}+\dfrac C{(n+2)!}$$

$n=-2\implies C=(-2)(-2+1)^2=-2$

Comparing the coefficients of $n^2,$ $$2=3+A\iff A=-1$$

Comparing the coefficients of $n,$ $$1=2+3A+B\iff B=1-2-3A=2$$

Interestingly, this leads to a Telescoping series,

otherwise, we had to use $$e^y=\sum_{r=0}^\infty\dfrac{y^r}{r!}$$

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Hint:  starting from OP's simplification, the following form will telescope nicely:

$$ \begin{align} \frac{n(n+1)^2}{ (n+2)!} &= \frac{(n+2)(n+1)n - (n+2)(n+1) + 2(n+2) -2}{(n+2)!} \\[5px] &= \frac{1}{(n-1)!} -\frac{1}{n!} + \frac{2}{(n+1)!} - \frac{2}{(n+2)!} \end{align} $$

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  • $\begingroup$ Both answer are great and correct. Yours is more direct but the other has more details. $\endgroup$
    – chanp
    Feb 6, 2018 at 6:55
  • $\begingroup$ @chanp The takeaway from either answer (+1 to @ lab btw) is that any polynomial can be written in terms of falling factorials, which is what you'd be looking for in a case like this in order to derive something resembling a partial fraction decomposition (with a chance for telescoping, though that's not guaranteed, and in fact rarely happens unless the problem was carefully set up so that it does). $\endgroup$
    – dxiv
    Feb 6, 2018 at 7:16

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