What is 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.

Answer 2

HINT

We have that

$$\sigma(n)=\sum_{i\neq j} \frac{1}{i^n j^n}=\sum_{i} \frac{1}{i^n}\sum_{j} \frac{1}{ j^n}-\sum_{k} \frac{1}{k^{2n}}$$