# Asymptotic formula for $\sum_{n\leq x}\sigma(n)$ knowing $\sum_{n\leq x}\frac{\sigma(n)}{n}$

Let $$\sigma(n):=\sum_{d|n}d$$ be the sum of all divisors of $$n$$. Find the asymptotic formula for $$\sum_{n\leq x}\frac{\sigma(n)}{n}$$ and use it to find the one for $$\sum_{n\leq x}\sigma(n)$$.

Here is my attempt: \begin{align*} \sum_{n\leq x}\frac{\sigma(n)}{n}=\sum_{n\leq x}\frac{1}{n}\sum_{d|n}\frac{n}{d} &=\sum_{n\leq x}\sum_{d|n}\frac{1}{d}\\ &=\sum_{d\leq x}\sum_{k\leq \frac{x}{d}}\frac{1}{d}\\ &=\sum_{d\leq x}\frac{1}{d}\sum_{k\leq \frac{x}{d}}1\\ &=\sum_{d\leq x}\frac{1}{d}\left\lfloor{\frac{x}{d}}\right\rfloor \end{align*} Since $$\left\lfloor\frac{x}{d}\right\rfloor=\frac{x}{d}+O(1)$$, we have: \begin{align*} \sum_{n\leq x}\frac{\sigma(n)}{n}&=\sum_{d\leq x}\left(\frac{x}{d^2}+O\left(\frac{1}{d}\right)\right)\\ &=x\underbrace{\sum_{d\leq x}\frac{1}{d^2}}_{=\frac{\pi^2}{6}+O\left(\frac{1}{x}\right)}+\underbrace{O\left(\sum_{d\leq x}\frac{1}{d}\right)}_{=O(\log(x))}\\ &=\frac{\pi^2}{6}x+\underbrace{O(1)+O(\log(x))}_{=O(\log(x))} \end{align*} Now to estimate $$\sum_{n\leq x}\sigma(n)$$, I thought about using Abel's summation: $$\sum_{n\leq x} \frac{\sigma(n)}{n}=\frac{1}{x}\left(\sum_{n\leq x}\sigma(n)\right)+\int_1^x\frac{1}{t^2}\left(\sum_{n\leq t}\sigma(n)\right)dt$$ I suppose that's a reasonable way, exept that I don't know what to do with $$\sum_{n\leq t}\sigma(n)$$.

Any suggestions?

• Use partial summation the other way around: start with $\sum_{n \leq x} \sigma(n)$ and relate it to $x \sum_{n \leq x} \frac{\sigma(n)}{n}$. – Peter Humphries Oct 15 '18 at 9:25

As demonstrated in my first attempt, $$\sum_{n\leq x}\frac{\sigma(n)}{n}=\frac{\pi^2}{6}x+O(\log x)$$. Now the trick is to write $$\sigma(n)=\frac{\sigma(n)}{n}\cdot n$$ and use Abel's summation: \begin{align*} \sum_{n\leq x}\sigma(n)&=\sum_{n\leq x}\frac{\sigma(n)}{n}\cdot n\\ &=x\underbrace{\left(\sum_{n\leq x}\frac{\sigma(n)}{n}\right)}_{=\frac{\pi^2}{6}x+O(\log x)}-\int_1^x\underbrace{\left(\sum_{n\leq t}\frac{\sigma(n)}{n}\right)}_{=\frac{\pi^2}{6}t+O(\log t)}dt\\ &=\frac{\pi^2}{6}x^2+O(x\log x)-\frac{\pi^2}{6}\int_1^xt\,dt+O\left(\int_1^x\log (t)dt\right)\\ \end{align*} Since $$\int_1^xt\,dt=\frac{x^2}{2}-\frac{1}{2}$$ and $$\int_1^x\log(t)dt=t\log t-t|_1^x=x\log x-x+1$$, we conclude: \begin{align*} \sum_{n\leq x}\sigma(n)&=\frac{\pi^2}{6}x^2+O(x\log x)-\frac{\pi^2}{12}x^2+\underbrace{\frac{\pi^2}{12}+O(x\log x-x+1)}_{O(x\log x)}\\ &=\frac{\pi^2}{12}x^2+O(x\log x) \end{align*}
• I believe using the exact same techniques as above we can arrive at $\sum_{n\le x}\sigma_k(n) \approx \frac{\zeta(k+1) }{k+1}x^{k+1}$. I am struggling to keep track of the error term precisely but that may be a project for another day (or some other interested on-looker). – Mason Oct 27 '18 at 22:27
• I figured out the error term! It should be on the order of $x^k$. Hooray! I am providing a link to this argument. – Mason Nov 5 '18 at 16:30