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

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
$$ \frac{1}{n} \sum_{k=1}^n f(n) \sim \frac{15}{\pi^2} +
\frac{6}{\pi^2}\zeta\left(\frac{1}{2}\right) n^{-1/2}.$$
