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The sum again is $$\sum_{n=1}^{\infty}\frac{1}{n!(n+2)}$$ It looks like it should be amenable to some modification to get towards the exponential power series or something close to it, But I really can't get anything to pop out.

Wolfram alpha says it sums to 1/2 and my friend verified this by partial sums but I am curious if there is a slick way to evaluate this, maybe something like partial fractions that can deal with the factorial or a product rule type thing?

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$$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$$ $$xe^x=\sum_{n=0}^{\infty}\frac{x^{n+1}}{n!}$$ $$\int_{0}^{1}xe^xdx=\int_{0}^{1}\sum_{n=0}^{\infty}\frac{x^{n+1}}{n!}dx$$ $$1=\sum_{n=0}^{\infty}\frac{1}{n!(n+2)}$$ $$1=\frac{1}{2}+\sum_{n=1}^{\infty}\frac{1}{n!(n+2)}$$

so $$\sum_{n=1}^{\infty}\frac{1}{n!(n+2)}=\frac{1}{2}$$

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    $\begingroup$ How is this answer different from the one posted one hour earlier by @stevenstadnicki ? $\endgroup$ – Mark Viola Sep 12 '16 at 1:40
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    $\begingroup$ @Dr.MV The reason I accepted it instead of the also very helpful ones posted by you and steven is the care taken with the definite integration. I think it is easier to avoid errors with indefinite integration. $\endgroup$ – qbert Sep 13 '16 at 13:25
  • $\begingroup$ FWIW, I never suggested specifically definite or indefinite integration; the reason mine is spelled out in less detail is that it was meant to be closer to a hint or outline than a complete top-to-bottom answer. $\endgroup$ – Steven Stadnicki Sep 26 '16 at 18:37
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HINT:

$$\frac{1}{n!(n+2)}=\frac{1}{(n+1)!}-\frac{1}{(n+2)!}$$

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  • $\begingroup$ I can definitely take it from here, thank you! $\endgroup$ – qbert Sep 11 '16 at 17:26
  • $\begingroup$ You're welcome! My pleasure. -Mark $\endgroup$ – Mark Viola Sep 11 '16 at 17:26
  • $\begingroup$ @DonAntonio Thank you my friend!! $\endgroup$ – Mark Viola Sep 11 '16 at 17:33
  • $\begingroup$ @E.H.E Very much appreciative. $\endgroup$ – Mark Viola Sep 11 '16 at 18:04
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Work via generating functions: $e^x=\sum_{n=0}^\infty\dfrac{x^n}{n!}$, so $xe^x=\sum_{n=0}^\infty\dfrac{x^{n+1}}{n!}$. Now integrate this with respect to $x$; you should find that the RHS becomes $\sum_{n=0}^\infty\dfrac{x^{n+2}}{(n+2)n!}$ (and the LHS is easily integrable by parts). Finally, set $x=1$ (and note that your sum starts at $n=1$ and not $n=0$, so there's a term from this series that you're missing).

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  • $\begingroup$ fantastic, I should have seen that. Thank you! $\endgroup$ – qbert Sep 11 '16 at 17:27
  • $\begingroup$ Very nice idea. +1 $\endgroup$ – DonAntonio Sep 11 '16 at 17:47

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