question about equality series containing hypergeometric term and a simple term 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?
 A: 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}
A: 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 $n$.
