Identity for $\zeta(k- 1/2) \zeta(2k -1) / \zeta(4k -2)$?

Question

Is there a nice identity known for $$\frac{\zeta(k- \tfrac{1}{2}) \zeta(2k -1)}{\zeta(4k -2)}?$$ (I'm dealing with half-integral $k$.) Equally, an identity for $$\frac{\zeta(s) \zeta(2s)}{\zeta(4s)}$$ would do ;)

Answer 1

Let $$F(s) = \frac{\zeta(s)\zeta(2s)}{\zeta(4s)}.$$ Then clearly the Euler product of $F(s)$ is $$F(s) = \prod_p \frac{\frac{1}{1-1/p^s}\frac{1}{1-1/p^{2s}}}{\frac{1}{1-1/p^{4s}}}= \prod_p \left( 1 + \frac{1}{p^s} + \frac{2}{p^{2s}} + \frac{2}{p^{3s}} + \frac{2}{p^{4s}} + \frac{2}{p^{5s}} + \cdots\right).$$ Now introduce $$ f(n) = \prod_{p^2|n} 2.$$ It follows that $$ F(s) = \sum_{n\ge 1} \frac{f(n)}{n^s}.$$ We can use this e.g. to study the average order of $f(n)$, given by $$ \frac{1}{n} \sum_{k=1}^n f(n).$$ The function $F(s)$ has a simple pole at $s=1$ and the Wiener-Ikehara-Theorem applies. The residue is $$\operatorname{Res}_{s=1} F(s) = \frac{15}{\pi^2}$$ so that finally $$ \frac{1}{n} \sum_{k=1}^n f(n) \sim \frac{15}{\pi^2}.$$ In fact I would conjecture that we can do better and we ought to have $$ \frac{1}{n} \sum_{k=1}^n f(n) \sim \frac{15}{\pi^2} + \frac{6}{\pi^2}\zeta\left(\frac{1}{2}\right) n^{-1/2}.$$