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This may look like homework, but it is not. I've found this identity (using Mathematica):

$$ {}_3F_2 \left( \matrix{1,1,1 \\ 2, e} ; 1 \right) = (e-1) \psi^{\prime}(e-1), $$

valid for $e$ with $\mathcal{R}(e)>0$, where ${}_3F_2$ is the Generalized Hypergeometric Function (as in here) and $\psi^{\prime}$ is the trigamma function (definition here).

It's also in Wolfram's site: http://functions.wolfram.com/07.27.03.0083.01

The problem is... I've no idea how to prove it. I've tried using the definition of the Pochhammer symbol and some simplifications to get this:

$$ {}_3F_2 \left( \matrix{1,1,1 \\ 2, e} ; 1 \right) = \sum_{k=0}^{\infty} \frac{\Gamma(k+1)\Gamma(e)}{(k+1)\Gamma(e+k)}, $$

but it's not even close to the series for trigamma function:

$$ (e-1) \psi^{\prime}(e-1) = (e-1) \sum_{n=0}^{\infty} \frac{1}{(e-1+n)^2}. $$

Any help/tips/references are appreciated.

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Euler's integral transformation at your 1st link allows to write the left side as an integral of an $_2F_1$ function \begin{align} _3F_2\left[\begin{array}{c}1,1,1\\2,e\end{array};1\right]=\frac{\Gamma(e)}{\Gamma(e-1)}\int_0^1 (1-t)^{e-2}{}_2F_1\left[\begin{array}{c}1,1\\2\end{array};t\right]dt, \end{align} which can be itself expressed via elementary functions: $$ {}_2F_1\left[\begin{array}{c}1,1\\2\end{array};t\right]=-\frac{\ln(1-t)}{t}.$$ (The last identity can be easily derived using series expansions of both sides). So $$ _3F_2\left[\begin{array}{c}1,1,1\\2,e\end{array};1\right]=-(e-1)\int_0^1 (1-t)^{e-2}\frac{\ln(1-t)}{t}dt.$$ Note that the integral on the right is perfectly well-defined at $t=0$, since $$\displaystyle\frac{\ln(1-t)}{t}=-1+\frac{t}{2}-\frac{t^2}{3}+\ldots$$ Moreover, this is nothing but the standard integral representation for $-\psi'(e-1)$; see e.g. the 2nd formula at your 2nd link.

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  • $\begingroup$ Thanks! I'll see if I can go on from here. $\endgroup$ – Ferdinand.kraft May 3 '13 at 13:10

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