This sum equal to? $\sigma(n)=\sum_{i\neq j} \frac{1}{i^n j^n}$

Question

I have recently come across the following sum, taken over all positive integers $i$ and $j$ such that $i \neq j$:

$$
\sigma(n)=\sum_{i\neq j} \frac{1}{i^n j^n},
$$

where $n$ is a positive integer greater than $1$.

Can this somehow be written in terms of the Riemann Zeta function? Is there already a zeta function of this kind?

As a corollary question, what about the sum

$$
\sigma(n)=\sum_{i\neq j} \frac{1}{i^n j^{n-1}}?
$$

Answer 1

Your first sum equals $$ \sum_{i,j}\frac1{i^nj^n}-\sum_{i=j}\frac1{i^nj^n}=\zeta(n)^2-\zeta(2n).$$

The same trick works for your the second sum.