Let $\psi_q(z)$ be the q-DiGamma function defined for a real variable $\Re(z)>0$ as $$\psi_q(z)=\frac{1}{\Gamma_q(z)}\frac{\partial}{\partial z} (\Gamma_q(z))$$ where $\Gamma_q(z)$ is the q-Gamma function defined as $$\Gamma_q(z)=(1-q)^{1-z}\prod_{n=0}^{\infty}\frac{1-q^{n+1}}{1-q^{n+z}}$$
Question I am looking for a closed form (to the extent possible) for $$\sum_{n=1}^\infty \psi_{e^{-2\pi}}(n+1)\left(\frac{1}{n^3}-\frac{1}{(n+1)^3}\right)$$
Since I was answering this on Mathematica Stack Exchange I will finish posting my effort
gamma[z_] := Evaluate[(1 - q)^(1 - z)*
Product[(1 - q^(n + 1))/(1 - q^(n + z)), {n, 0, Infinity}]]
psi[z_] := Evaluate[D[1/(gamma[z]), z]*gamma[z]]
q = E^(-2 Pi);
Sum[psi[n + 1]*(1/n^3 - 1/(n + 1)^3), {n, 1, Infinity}]
Looking at the psi[n + 1] term
psiSimplified = Simplify[psi[n + 1]]
-QPolyGamma[0, -(Log[E^(-2 (1 + n) Pi)]/(2 Pi)), E^(-2 Pi)]
I'm not sure how Fred Hutch gets to the next step:
sum = Sum[QPolyGamma[n + 1, E^(-2*Pi)]*(1/n^3 - 1/(n + 1)^3), {n, 1, Infinity}]
but it doesn't result in a closed form.
N[sum]
0.00184998
q-gamma functionis a built-in symbol in Mathematica. Theq-digamma functionis also a built-in symbol in Mathematica. – Bob Hanlon Apr 10 '23 at 15:55