For positive integer $n$ the following value of a hypergeometric function


$$\_4F\_3(n,n,n,2n,1+n,1+n,1+n,-1)$$


based on the first few terms looks like


$$ R\_1(n) + R\_2(n) \pi^2$$


where $R\_{1,2}(n)$ are rational numbers. Is there a way to express $R\_{1,2}(n)$ in a closed form using “simple” operations like power, ratio, factorial, etc?



Closed form for ₄F₃(n,n,n,2n;1+n,1+n,1+n;−1)
Fiddling with Maple, I get: if $n$ is a positive real number, then
$$
{{\_4\mathrm F\_3}(n,n,n,2\,n;\,n+1,n+1,n+1;\,-1)}={\frac {{n}^{2}
\sqrt {\pi}\,\Gamma(n+1)\,\psi^{(1)}(n)}{{4}^{n}\,\Gamma \left( n
+{\frac{1}{2}}\right)}. }
$$
Here, $\psi^{(1)}$ is the [trigamma function](https://en.wikipedia.org/wiki/Trigamma_function),
$\psi^{(1)}(z) = \frac{d^2}{dz^2}\log(\Gamma(z))$.




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It seems, for positive integer $n$, we have that $\psi^{(1)}(n)$ has the form $\pi^2/6 - \text{rational}$.


Francois pointed out the recurrence, so that
$$
\psi^{(1)}(n) = \frac{\pi^2}{6}-\sum\_{k=1}^{n-1}\frac{1}{k^2}
$$


