Digamma function property

Question

I am trying to determine $\displaystyle\sum_{n=0}^{\infty}\dfrac{1}{(\alpha\cdot n+1)^2} $, to do this I will use the following property that I cannot prove, any ideas?

$$ \sum_{n=0}^\infty\frac1{(\alpha n+1)^2}=\frac1{\alpha^2}\psi'\left(\frac1\alpha\right) $$

I know some properties about the digamma function but almost none about its derivative, I need some idea to start since I can only think of deriving the definition $\psi(z)=\dfrac{\Gamma'(z)}{\Gamma(z)}$.

Answer 1

The $m$th derivative of the digamma function is the polygamma function of order $m$, $\psi^{(m)}(z)$. The polygamma function has the series representation $$\psi^{(m)}(z) = (-1)^{m+1}m! \sum_{n=0}^{\infty}\frac{1}{(z+n)^{m+1}},$$ so writing $(\alpha n+1)^2 = \alpha^2\left(n+\frac{1}{\alpha}\right)^2$ (and taking $m = 1$) gives your series.

This series expansion follows directly from the recurrence relation $$\psi^{(m)}(z+1) = \psi^{(m)}(z) + \frac{(-1)^m m!}{z^{m+1}},$$ which we obtain by differentiating ($m$ times) the corresponding recurrence relation for the digamma function, $$\psi(z+1) = \psi(z) + \frac{1}{z}$$

Answer 2

Maple evaluates these: $$ \psi'\left(1\right) = \frac{\pi^2}{6} \\ \psi'\left(\frac{1}{2}\right) = \frac{\pi^2}{2} \\ \psi'\left(\frac{1}{4}\right) = \pi^2 + 8\;\text{Catalan} $$ but no others among $\psi'(1/k)$ with $k$ up to $20$.