I am trying to prove the following identity:
$$ \sum_{n=1}^\infty \frac{\sigma_a(n)\sigma_b(n)}{n^s} = \frac{\zeta(s)\zeta(s-a)\zeta(s-b)\zeta(s-a-b)}{\zeta(2s-a-b)}, \qquad (\Re s > 1+\Re (a+b), \Re (a) \geq 0,\Re(b) \geq 0) $$ If we assume that the sum on the left is convergent, this can be proved by first assuming $\Re(a),\Re(b) \neq 0$ and then taking a limit from the right.
However, how to prove that the sum on the left is convergent under these conditions without using more advanced results such as Gronwall's theorem? It suffices to show that $\sigma_a(n)$ grows asymptotically slower than $n^{\Re(a)}$, since from which we can bound the left side by a p-series.