I was looking at this series $$ f(z)=\sum_{k=1}^{\infty}\frac{1}{(k+\frac{1}{k})^{z}} $$ and wondering if it is somehow realted to the Riemann's zeta function $$ \zeta(z)=\sum_{k=1}^{\infty}\frac{1}{k^{z}} $$ Does anyone knoes if there is a relation between $f(z)$ and $\zeta(z)$? The kind of relation I'm looking for is something like $f(z)=g(z)\zeta(z)$. I'd appreciate some references too.
Thanks in advance.
EDIT:
Meanwhile I was able to derive the following relation between $f(s)$ and $\zeta(s)$. Enjoy!
$$ f(s)=\sum_{k=1}^{\infty} \frac{1}{\left(k+\frac{1}{k}\right)^{s}}=\sum_{k=0}^{\infty}(-1)^{k} \frac{s^{(k)}}{k !} \zeta(s+2 k) $$ where $s^{(k)}=s(s+1) \cdots(s+k-1)$ is the rising factorial.
Now we can express $\zeta(s+n)$ as $$ \zeta(s+n)=\zeta(s)\times\prod_{p\in\mathbb{P}}\frac{p^{s+n}-p^{n}}{p^{s+n}-1} $$ and for this particular case we have $$ \boxed{ \;\;\;\; \sum_{k=1}^{\infty}\frac{1}{(k+\frac{1}{k})^{s}}=\zeta(s)\left(1+\sum_{k=1}^{\infty}(-1)^{k}\frac{s^{(k)}}{k!}\prod_{p\in\mathbb{P}}\frac{p^{s+2k}-p^{2k}}{p^{s+2k}-1} \right ) \;\;\;\;} $$ as I expected.
For large $k$, $(k+1/k)^z =k^z(1+1/k^2)^z \approx k^z(1+z/k^2) = k^z+zk^{z-2} $ so, for $k^2 > z$,
$\begin{array}\\ \frac1{k^z}-\frac1{(k+1/k)^z} &\approx \frac1{k^z}-\frac1{k^z+zk^{z-2}}\\ &=\frac{(k^z+zk^{z-2})-k^z}{k^z(k^z+zk^{z-2})}\\ &=\frac{zk^{z-2}}{k^{2z}(1+zk^{-2})}\\ &\approx\frac{z}{k^{z+2}}(1-zk^{-2}+O(zk^{-4}))\\ &\approx \frac{z}{k^{z+2}}-\frac{z^2}{k^{z+4}}\\ \end{array} $
This sort of indicates that it might be true that $f(z) \sim \zeta(z)-z\zeta(z+2)+z^2\zeta(z+4) $ and that there is an expansion (maybe) that continues this.
As to how accurate this is, I have no idea.