Is $f(z)=\sum_{k=1}^{\infty}\frac{1}{(k+\frac{1}{k})^{z}}$ somehow related to Riemann's zeta function?

Question

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.

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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.

Answer 1

$$(n+n^{-1})^{-s} = n^{-s} (1+n^{-2})^{-s}= n^{-s} \sum_{k=0}^\infty {-s \choose k} n^{-2k}$$ so that for $\Re(s) > 1$ $$\sum_{n=1}^\infty (n+n^{-1})^{-s} = 2^{-s}+\sum_{k=0}^\infty {-s \choose k} (\zeta(s+2k)-1)$$

The RHS extends meromorphically to $\Bbb{C}$ with simple poles at negative integers.

Answer 2

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.