<s>[INST] $\pi$ in terms of polygamma The computer found this, but couldn't prove it.
Let $\psi(n,x)$ denote the polygamma function.
With precision 500 decimal digits we have:
$$ \pi^2 = \frac{1}{4}(15 \psi(1, \frac13) - 3 \psi(1, \frac16)) $$
Is it true?
In machine readable form:
 pi^2 == 1/4*(15*psi(1, 1/3) - 3*psi(1, 1/6))

 [/INST] Note that 
$$ 
\psi(m,x) =(-1)^{m+1} m! \sum_{k=0}^{\infty} \frac{1}{(x+k)^{m+1}}.
$$ 
Therefore 
$$ 
\psi(m,1/6) = (-1)^{m+1} m! \sum_{k=0}^{\infty} \frac{1}{(k+1/6)^{m+1}} =(-1)^{m+1} m! 6^{m+1} \sum_{n\equiv 1 \mod 6} \frac{1}{n^{m+1}}. 
$$ 
Writing the condition $n\equiv 1 \mod 6$ as $n\equiv 1 \mod 3$ but not $4 \mod 6$, the above is 
\begin{align*}
&(-1)^{m+1} m! 6^{m+1} \Big( \sum_{n\equiv 1 \mod 3} \frac{1}{n^{m+1}} - \frac{1}{2^{m+1}} \sum_{n\equiv 2 \mod 3} \frac{1}{n^{m+1}}\Big)\\
&= 2^{m+1} \psi(m,1/3)-\psi(m,2/3). 
\end{align*}
We also have 
$$
\psi(m,1/3) +\psi(m,2/3) = (-1)^{m+1} m! 3^{m+1} \sum_{n \not\equiv 0\mod 3} \frac{1}{n^{m+1}} = (-1)^{m+1} m! (3^{m+1} -1) \zeta(m+1). 
$$
From these two relations, clearly we have a linear relation connecting $\psi(m,1/6)$, $\psi(m,1/3)$ and $\zeta(m+1)$: namely, 
$$ 
\psi(m,1/6) = (2^{m+1}+1) \psi(m,1/3)+(-1)^m m! (3^{m+1}-1) \zeta(m+1).
$$
</s>