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I am having some difficulties to find the solution for this sum of fractions with factorials in denominator. Any ideas how to solve this?
$$\sum_{i=1}^{2016}\frac{1}{i!+(i+1)!+(i+2)!} =\frac{1}{1!+2!+3!}+\frac{1}{2!+3!+4!}+...+\frac{1}{2016!+2017!+2018!}=?$$

I have used the following formula as a one way of solving it, but didn't get good results. $$\frac{1}{n!+(n+1)!+(n+2)!}=\frac{1}{n!(n+2)^2}$$

It turns out that we can solve/represent the solution using hypergeometric functions. Can anyone suggest the elementary way of solving/simplifying it?

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  • $\begingroup$ Hi! Welcome to MSE. What kind of simplifications have you been trying? $\endgroup$ – Arnaud Mortier Jan 27 '18 at 15:41
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    $\begingroup$ Hi there, thanks! Edited the question. $\endgroup$ – Mirzodaler Jan 27 '18 at 15:53
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Doesn't seem like it's anything known.

Wolfy says that the sum is about 0.1503796770046413405002786271034306597823, and the inverse symbolic calculator doesn't give anything useful.

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Maple gives $$ \sum_{i=1}^n \frac{1}{i!(i+2)^2}=\frac 1 9\,{\mbox{$_3$F$_3$}(1,3,3;\,2,4,4;\,1)}-{\frac { {\mbox{$_3$F$_3$}(1,n+3,n+3;\,n+4,n+4,n+2;\,1)}}{ \left( n+1 \right) ! \, \left( n+3 \right) ^{2}}}, $$ so $$ \sum_{i=1}^{2016} \frac{1}{i!(i+2)^2} \approx \frac 1 9\,{\mbox{$_3$F$_3$}(1,3,3;\,2,4,4;\,1)}. $$

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  • $\begingroup$ Thanks for that. Can you please explain the 3F3 function there, it is something new for me. What is formula for that or any other links to understand it. $\endgroup$ – Mirzodaler Jan 28 '18 at 5:11
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    $\begingroup$ see Hypergeometric Function mathworld.wolfram.com/HypergeometricFunction.html $\endgroup$ – Leox Jan 28 '18 at 9:11

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