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?



question about equality series containing hypergeometric term and a simple term
By equation (1.22) in 


*Nørlund, Niels Erik*, [**Hypergeometric functions**](http://dx.doi.org/10.1007/BF02392494), Acta Math. 94, 289-349 (1955). [ZBL0067.29402](https://zbmath.org/?q=an:0067.29402).


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}


