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Evaluate the following sum: $$\dfrac{1}{1!+2!+3!}+\dfrac{1}{2!+3!+4!}+\dots + \dfrac{1}{2016!+2017!+2018!}$$

I was trying to rewrite the general term as: $$\frac{1}{n!+(n+1)!+(n+2)!}=\frac{1}{n!(n+2)^2}$$

However, this did not give any essential improvements.

RFZ
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  • I believe the general term can be rewritten as $\frac{1}{2*n!(n+2)}$ instead. – Szeto Jan 22 '18 at 08:02
  • @Szeto, I don't think so. Could you demonstrate it? – RFZ Jan 22 '18 at 08:03
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    @Szeto $$n!+(n+1)! + (n+2)! = n!(1 + (n+1) + (n+1)(n+2))\= n!(1\cdot (n+2) + (n+1)(n+2)) = n!(n+2)(1 + n+1)$$ – Arthur Jan 22 '18 at 08:03
  • @Arthur Oh no I’ve made a silly mistake when doing math in my mind. Shouldn’t have been that careless. – Szeto Jan 22 '18 at 08:05
  • The sum of the infinite series has no explicit form. The sum of the finite series you are asking about might be even worse. What makes you think a simple formula exists? – Did Jan 22 '18 at 08:17
  • @Did, i was thinking that one can use telescope method but I guess that its finite sum is difficult to calculate. – RFZ Jan 22 '18 at 08:25
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    You tagged this as (contest-maths). Which contest does this come from? – Did Jan 22 '18 at 08:31
  • @Did, No it is not from contest math. I was thinking that it is look like problems which one can encounter in olympiads. The last time I participate in contests was in high school about 7-8 years ago :) – RFZ Jan 22 '18 at 08:32
  • Why downvote? :( – RFZ Jan 22 '18 at 08:56
  • "Why downvote?" For once, I share the OP's puzzlement about a downvote... Having said that, I still think that it is generally poor form to post a comment enquiring about the reasons of a downvote as soon as one receives one (and I am still waiting to be witnessing the same reaction to upvotes...). – Did Jan 22 '18 at 16:28

2 Answers2

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Using the exponential integral function $\text{Ei}$, one can show that $$\sum_{n=1}^\infty\frac{x^{n+2}}{n!(n+2)^2}=e^x-\frac14x^2-1+\ln x-\text{Ei}(x)+\gamma$$ hence $$\sum_{n=1}^\infty\frac{1}{n!(n+2)^2}=e-\frac54-\text{Ei}(1)+\gamma$$ Since there is no expression of $\text{Ei}(1)$ except as the sum of a series quite related to the infinite series corresponding to the series in the question, this mainly shows the question has no satisfying answer.

Did
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$$\sum_{k = 0}^{+\infty} \frac{1}{n! + (n+1)! + (n+2)!} = -\text{Ei}(1)+\gamma +e-1$$

Where $Ei$ is the Exponential Integral special function, and $\gamma$ is the Euler-Mascheroni constant.

More generally we have:

$$\sum_{k = 0}^{N} \frac{1}{n! + (n+1)! + (n+2)!} = $$

$$ = \frac{(-\text{Ei}(1)+\gamma +e-1) ((N+1)!+(N+2)!+(N+3)!)-\, _3F_3(1,N+3,N+3;N+2,N+4,N+4;1)}{(N+1)!+(N+2)!+(N+3)!}$$

Where you are facing the Hypergeometric special function.

Enrico M.
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    ...Which is just a fancy way of rewriting the original sum. – Did Jan 22 '18 at 08:29
  • For the $N$ case, it is. But you can do numbers with that, by knowing some Hypergeometric special function values! – Enrico M. Jan 22 '18 at 12:59
  • Is one of these known special values of hypergeometric functions providing an answer in the present case? – Did Jan 22 '18 at 16:25