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I need a help about the following: Maple gave that the following equality is true for n =1,2,3,4,5, $$ \sum_{h=0}^{\infty}\binom{n+h}{n}{_3}F_2\left( \substack{-h,n+1,n+1\\ 1,1}; x\right)= \frac{1}{x^{n+1}}.$$

Note that ${_3}F_2\left( \substack{-h,n+1,n+1\\ 1,1}; x\right)$ is a polynomial of degree $h$.

I would like to know why that is true? Maple fails to compute for $n=6$ in my computer, but is it true for all integer $n \geq 6$?
Where can i find such series involving hypergeometric term?

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  • $\begingroup$ For $\ n=0\ $ the sum is $\ \sum_{h=0}^\infty (1-x)^h\ $ which converges to $1/x$ iff $\ 0<x<2.\ $ Similar results hold for $\ n>0.$ $\endgroup$ – Somos Apr 19 at 5:44
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By equation (1.22) in

Nørlund, Niels Erik, Hypergeometric functions, Acta Math. 94, 289-349 (1955). ZBL0067.29402.

the sum in question equals $$\lim_{z\to 1} \frac{1}{(1-z)^{n+1}}\ {_3}F_2\left({n+1,n+1,n+1\atop 1,1}; \frac{xz}{z-1}\right).$$

It appears (although I did not check carefully) that $${_3}F_2\left({n+1,n+1,n+1\atop 1,1}; \frac1t\right) = (-t)^{n+1} \sum_{k\geq 0} \binom{n+k}{k}^3 t^k.$$ Hence, \begin{split} &\lim_{z\to 1} \frac{1}{(1-z)^{n+1}}\ {_3}F_2\left({n+1,n+1,n+1\atop 1,1}; \frac{xz}{z-1}\right) \\ &= \lim_{z\to 1} \frac{1}{(xz)^{n+1}}\sum_{k\geq 0} \binom{n+k}{k}^3 \left(\frac{z-1}{xz}\right)^k \\ &= \frac{1}{x^{n+1}}. \end{split}

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  • $\begingroup$ thank you ( Mr Andras and Max) for your help, any one have an idea about the proof? $\endgroup$ – mamiladi Apr 19 at 11:55
  • $\begingroup$ @mamiladi: I've completed the proof. $\endgroup$ – Max Alekseyev Apr 19 at 12:09
  • $\begingroup$ thanks very much mr Max, it's correct $\endgroup$ – mamiladi Apr 19 at 12:45
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It holds as well for $n=6, 7, 8, 9, 10$.

Evidently Mathematica is using the same algorithm for reduction as Maple because it too fails to reduce this exactly whn $n \geq 6$.

The root cause may be that the convergence is rather slow: For example, at $n=6, x = \frac13$ the expression does not even stay positive until you sum to $h \geq 121$. It is plausible that both programs examine the behavior of the sum, and when they see it begin to converge make an ansatz as to the infinite sum value, and then manipulate the series to try to prove the ansatz. The problem may come when that first step does not lead to an apparent answer, and each program goes on to try alternative methods, which don't work.

By the way, the function does not appear to sum to $x^{-(n+1)}$ for fractional $n$.

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  • $\begingroup$ thanks mr Mark for your answer, thank you for cheking the equality for n=6,7,8,9,10. that equality interess me for small value of x ( 0<|x|<1), i would like understand the raison of the result = $ \displaystyle \frac{1}{x^{n+1}}$, if you don't mind can you please try to check it for $n=11,12$ thank you. if it's true for all $n$ and $ |x|<1 $ Isdo you think it's a new result? $\endgroup$ – mamiladi Apr 18 at 23:45
  • $\begingroup$ It appears to work for $n=11$ and $n=12$, and $n=13$ but for smaller $N$ values I was able to verify an array of $x$ values that was sufficient to show that the polynomial multiplying $x^{-(n+1)}$ is just $1$ out to quite a few terms, whereas for $n=12$ it is impractical to use such a large spectrum of $x$ values so only a modest sample were verified. I think a straight proof of the overall theorem may be tractable specifically when $x=\frac12$ but I get bogged down when trying to prove that case. $\endgroup$ – Mark Fischler Apr 18 at 23:56
  • $\begingroup$ As to whether it is a new result, I would not be surprised to learn that Ramajudan scribbled it on a napkin somewhere while warming up for the real challenging stuff (LOL). $\endgroup$ – Mark Fischler Apr 19 at 0:00
  • $\begingroup$ thanks very much for your quick answer, i will try to find a proof.... $\endgroup$ – mamiladi Apr 19 at 0:09
  • $\begingroup$ i would like to ask you a question: did you try by mathmatica to find a recurrence relation to the quantity putted at the question involving binomial coefficient? $\endgroup$ – mamiladi Apr 19 at 0:17

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