For integers $a,b,k$, how would one evaluate (or find an upper/lower bound) to $$\sum_{n≤k}\frac{\sigma(n)^a}{n^b}$$ Where $\sigma(n)$ denotes the divisor sigma function?
$\bf{Edit:}$ More specifically, how would one evaluate $$\sum_{n≤k}\frac{(\sigma(n)-n)^s}{n^{s-1}}$$ for some integer $s>2$?
It is well-known that: $$ \sum_{n\leq x}\sigma(n) = \frac{\pi^2}{12} x^2 + O(x)\tag{1} $$ and the average order of $\sigma(n)^a$ can be derived from Euler's product and tauberian theorems.
For instance
$$\begin{eqnarray*}\sum_{n\geq 1}\frac{\sigma(n)^2}{n^s}&=&\prod_{p}\left(1+\frac{(p+1)^2}{p^s}+\frac{(1+p+p^2)^2}{p^{2s}}+\ldots\right)\\&=&\prod_p\frac{1-\frac{1}{p^{2s-2}}}{\left(1-\frac{1}{p^s}\right)\left(1-\frac{1}{p^{s-1}}\right)^2\left(1-\frac{1}{p^{s-2}}\right)}\\&=&\frac{\zeta(s)\zeta(s-1)^2\zeta(s-2)}{\zeta(2s-2)}\tag{2}\end{eqnarray*}$$ leads to $$ \sum_{n\leq x}\sigma(n)^2 = \frac{5\zeta(3)}{6} x^3+O(x^2)\tag{3} $$ and $\sum_{n\leq x}\frac{\sigma(n)^2}{n^b}$ can be estimated through summation by parts. Of course the behaviour of $\sum_{n\leq x}\frac{\sigma(n)^a}{n^b}$ strongly depends on $b>(a+1)$ (leading to a convergent series) $b=(a+1)$ (leading to a slowly divergent series) or $b<(a+1)$ (leading to a polynomial bound).