We know from the digamma function
$$ \Psi (z+1)= -\gamma -\sum_{k=1}^{\infty}\zeta(k+1)(-z)^{k} $$
My question is if there is a similar formula for
$$ f(a)+ \sum_{k=1}^{\infty}\zeta_{H}(k+1,a)(-z)^{k}\quad?$$
Using Ramanujan's master theorem i would say that i should take the mellin inverse of
$$ \frac{\pi}{\sin(\pi s)}\zeta _{H} (1-s,a) $$
The considered generating function admits a closed form.
Proposition. Let $a$ be a complex number such that $\Re a>0$. Then
$$ \sum_{k=1}^{\infty}\zeta_{H}(k+1,a)(-z)^{k}=\psi(a)-\psi (a+z),\qquad |z|<|a|, $$ where $\psi:=\Gamma'/\Gamma$.
Proof. Let $a$ be a complex number such that $\Re a>0$ and let $z$ be a complex number satisfying $|z|<|a|$.
We have$$ \begin{align} \sum_{k=1}^{\infty}\zeta_{H}(k+1,a)(-z)^{k}&=\sum_{k=1}^\infty \left(\sum_{n=0}^\infty\frac1{(n+a)^{k+1}}\right)(-z)^{k}\\\\ &=\sum_{n=0}^\infty \frac1{(n+a)}\sum_{k=1}^\infty\left(-\frac{z}{(n+a)}\right)^{k}\\\\ &=-\sum_{n=0}^\infty \frac{z}{(n+a)(n+a+z)}\\\\ &=\sum_{n=1}^\infty \frac{z}{n(n+a-1)}-\sum_{n=1}^\infty \frac{z}{n(n+a+z-1)}\\\\ &=\psi(a)-\psi (a+z) \end{align} $$ where we have used a standard series representation (6.3.16) for the digamma function.