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

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.

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 $$\boxed{ \;\;\;\; \sum_{k=1}^{\infty}\frac{1}{(k+\frac{1}{k})^{z}}=\zeta(s)\left(1+\sum_{k=1}^{\infty}(-1)^{k}\frac{s^{(s)}}{k!}\prod_{p\in\mathbb{P}}\frac{p^{s+2k}-p^{2k}}{p^{s+2k}-1} \right ) \;\;\;\;}$$ as I expected.

• No, there is no meaningful connection between the two, other than their superficial similarity in both definition and numerical value. If anything, I would search for a connection to the polygamma function (a generalization of harmonic numbers), as well as trigonometric and hyperbolic functions. – Lucian Apr 20 '15 at 14:28
• $g(x)=x$ and $h(x)=x+\dfrac1x$ are two wholly different geometric shapes. The former is a straight line, the latter a hyperbola. – Lucian Apr 20 '15 at 14:35

$$(n+n^{-1})^{-s} = n^{-s} (1+n^{-2})^{-s}= n^{-s} \sum_{k=0}^\infty {-s \choose k} n^{-2k}$$ 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.
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.