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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.

Ѕᴀᴀᴅ
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Tom W.
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    Oh come on. Close voter, please just leave a comment letting the OP know what you think is wrong with his question and give him a chance to improve rather than just close voting right off the bat without saying anything. Imagine we all did that, questions would get closed with new people stumped as to what they did wrong –  Mar 16 '18 at 00:43
  • I was finally able to solve this. Due to Ramanujan, the leading coefficient is $5\zeta(3)x^3/6$ – Tom W. Mar 16 '18 at 13:25
  • Aye, that's pretty cool! Also not sure why this question has been closed, it's genuinely a good question. –  Mar 16 '18 at 14:23

2 Answers2

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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.

Jack D'Aurizio
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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 $x^3$...

qwr
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  • I appreciate the response as it's an interesting theorem I have never heard of before. I tried to manipulate this, but I was unable to arrive at the result described before. For example $\sum _{n \leq 1000}{\sigma_1(n)^2} = 1001531061$. This is very close to 1000^3, and while I can arrive that the sum is of some cubic order, my main problem is proving that it is close to $x^3$. If you have any more suggestions, that would be most appreciated. – Tom W. Mar 16 '18 at 02:43
  • @TomW. I've found a much more elementary result that is a good reference to the ideas used – qwr Mar 16 '18 at 03:01
  • Thanks for the help. I tried to do something akin to finding the difference between $\sum_{n \leq x} {\sigma_1(x)^2}$ and $\sum_{n \leq x} {\sigma_1(x)}*\sum_{n \leq x} {\sigma_1(x)}$, but I couldn't end up simplifying it to just $x^3$. Another relation I came up with was $\sigma_1(x)^2 \approx 2\sigma_1(x^2)-\sigma_2(x)+c$ where $c \leq \sigma_2(x)$. That relation didn't seem to help much though. This problem has been troubling me for quite some time :) – Tom W. Mar 16 '18 at 03:13