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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 $e^x$.

Bryden C
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chanp
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2 Answers2

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

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

dxiv
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  • Both answer are great and correct. Yours is more direct but the other has more details. – chanp Feb 06 '18 at 06:55
  • @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). – dxiv Feb 06 '18 at 07:16
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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 $$e^y=\sum_{r=0}^\infty\dfrac{y^r}{r!}$$

lab bhattacharjee
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  • Got it. If it does not telescope then multiple of the exponential term will appear as a result. – chanp Feb 06 '18 at 06:57
  • @chanp, See also : https://math.stackexchange.com/questions/576976/evaluate-the-series-lim-limits-n-to-infty-sum-limits-i-1n-fracn22/576978#576978 and https://math.stackexchange.com/questions/581603/evaluate-frac13-frac14-frac12-frac15-frac13-dots and https://math.stackexchange.com/questions/845464/determine-the-value-of-the-series-sum-n-1-infty-frac1-n2-n – lab bhattacharjee Feb 06 '18 at 07:01