Figuring out zeros and divergence of a double series

Question

I’m trying to figure out an interesting double series, but can’t figure out how to represent it as a single series, and know little about double series. $$ \sum_{n=1}^∞ \sum_{m=1}^∞ \frac{\cos (y\ln(n/m))}{(n^x)(m^x)} $$ is the double series. I can see that the central diagonal, if one were to plot this out on a plane, is equal to the Riemann zeta function of 2x, as any number divided by itself is one, the natural log of one is zero, zero times any value of y is zero, and cosine of zero is one. I am trying to figure out for what values of y and x does this series diverge, where it converges, and where it equals zero.

Answer 1

For $x > 1$ this series is easily seen to be absolutely convergent to $C^2(x, y) + S^2(x, y)$, where $C(x, y) = \displaystyle\sum_{n = 1}^{\infty} \frac{\cos(y\ln n)}{n^x}$ and $S(x, y) = \displaystyle\sum_{n = 1}^{\infty} \frac{\sin(y\ln n)}{n^x}$. Essentially, the sum is $|\zeta(x+iy)|^2$, and it is known that for $x > 1$ there are no zeros.

If there is some form of conditional (non-absolute) convergence to be analysed, the question is what exactly this form is (for multiple series, this is important - there's no widespread conventions on that).