Assume that $\Re(z)>1$ and $x\in \mathbb R$ with $ x > -3$.


Is it possible to compute $$\sum\_{n=0}^{\infty} \sum\_{k=0}^{3^n-1} \frac1 { \left(3\times3^n+kx\right)^z }\ ?$$


I believe that this sum is related in some way to the Riemann zeta function, but I can not prove it.



Series $\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{3^n-1}\left(3\cdot3^n+kx\right)^{-z }$ with $\Re(z)>1$ and $ x > -3$ real
This is just a **formal computation** to relate your function with Hurwitz zeta-function
$$
\zeta(s,x)=\sum\_{k\ge0}\frac1{(x+k)^s}.
$$


I **assume** convergence and a correct domain of definition:


\begin{align}
\notag\sum\_{n\ge0}\sum\_{k=0}^{3^n-1}\frac1{\left(3^{n+1}+kx\right)^z}&=
\sum\_{n\ge0}\sum\_{k\ge0}\frac1{\left(3^{n+1}+kx\right)^z}
-\sum\_{n\ge0}\sum\_{k\ge3^n}\frac1{\left(3^{n+1}+kx\right)^z}\\
\notag &=\sum\_{n\ge0}\sum\_{k\ge0}\frac1{\left(3^{n+1}+kx\right)^z}
-\sum\_{n\ge0}\sum\_{k\ge0}\frac1{\left(3^{n+1}+(k+3^n)x\right)^z}\\
\notag &=\sum\_{n\ge0}\sum\_{k\ge0}\frac1{\left(3^{n+1}+kx\right)^z}
-\sum\_{n\ge0}\sum\_{k\ge0}\frac1{\left(3^{n}(x+3)+kx\right)^z}\\
\notag &=x^{-z}\sum\_{n\ge0}\sum\_{k\ge0}\frac1{\left(3^{n+1}/x+k\right)^z}
-x^{-z}\sum\_{n\ge0}\sum\_{k\ge0}\frac1{\left(3^{n}(x+3)/x+k\right)^z}\\
\notag &=x^{-z}\sum\_{n\ge0}\zeta\left(z,3^{n+1}/x\right)
-x^{-z}\sum\_{n\ge0}\zeta\left(z,3^{n}(x+3)/x\right).
\end{align}


