Average order of $\sigma_1(n)^2$

Question

I am working on trying to find the average order of $\sigma_1(n)^2$, and I so far have found the result $$\sum_{n \leq x} {\sigma_1(n)^2} \approx x^3.$$ A derivation of this has eluded me and I was wondering if anyone else had an idea on how to get this.

Answer 1

This is an approximate argument from Carl Pomerance's notes:

$$\frac 1 x \sum_{n \le x} \sigma (x) = \frac{\pi^2}{12}x + O(\log x)$$

So on average, $\sigma (x)$ is on the order of $x$, and we know the sum of squares is on the order of $x^3$...

Answer 2

One may consider that, by Euler's product $$ \sum_{n\geq 1}\frac{\sigma(n)^2}{n^s} = \prod_{p}\frac{p^{2s}(p^s+p)}{(p^s-1)(p^s-p)(p^s-p^2)} = \prod_p\frac{\left(1-\frac{1}{p^{2s-2}}\right)}{\left(1-\frac{1}{p^s}\right)\left(1-\frac{1}{p^{s-1}}\right)^2\left(1-\frac{1}{p^{s-2}}\right)}$$ for any $s$ with a sufficiently large real part (namely $\text{Re}(s)>3$). The RHS equals $$ \frac{\zeta(s)\,\zeta(s-1)\,\zeta(s-2)}{\zeta(2s-2)}=\frac{15\,\zeta(3)}{\pi^2(s-3)}+O(1)\quad\text{as }s\to 3^+ $$ hence Ramanujan's result $$ \sum_{n\leq x}\sigma(n)^2 \sim \frac{5\,\zeta(3)}{6} x^3$$ can be recovered from the previous lines through Hardy-Littlewood tauberian theorem.
Actually $\frac{5}{6}\zeta(3)$ is very close to $1$, and that is not entirely coincidental, as already discussed on MSE.