Given $[n, m, a] \in \Bbb R$, what is the partial sum formula for:
$$\sum_{x=1}^m\frac{H_{n,2x-1}}{a^x} = ???$$
Where $H_{x,y}$ is the generalized harmonic number. For context, while working on the a proof involving the polygamma function, I came across:
$$\sum_{x=1}^m\frac{\psi^{(2x-2)}(n+1)-\psi^{(2x-2)}(1)}{a^x (2x-2)!} =\ ???$$
Thanks to Wolfram Research's Functions Site I deduced that: $$\sum_{x=1}^m\frac{\psi^{(2x-2)}(n+1)-\psi^{(2x-2)}(1)}{a^x (2x-2)!} =\ \sum_{x=1}^m\frac{-[\zeta(2x-1)-H_{n,\ 2x-1}]-(-\zeta(2x-1))}{a^x}$$
Which of course is just: $$\sum_{x=1}^m\frac{H_{n,\ 2x-1}}{a^x} =\ ???$$
EDIT: I'm getting very close to an answer for this! The depth of the problem is actually quite shocking, to me at least!