Counting argument for $\sum \frac{\sigma}{n}$

Question

WTS $\sum_{n \leq x } \frac {\sigma(n)}{n} = \frac {\pi^2}{6}x + O(\log (x)) $

First we have that $\sum_{n\leq x} \frac{\sigma (n)}{n} = \sum_{n\leq x} \frac{1}{n} \sigma (n)= \sum_{n\leq x} \frac{1}{n} \sum_{d|n} d $ we now flip the sums; $\sum_{d\leq x} d \sum_{n \leq x, d|n} \frac{1}{n} =\sum_{d\leq x} \sum_{n \leq x, d|n} \frac{d}{n} $

We notice that $ d|n \iff \frac{n}{d}|n $ so we may replace $\frac {d}{n} $ with $\frac{\frac{n}{d}}{n} = \frac {1}{d}$ it follows that $\sum_{d\leq x} \sum_{n \leq x, d|n} \frac{d}{n} =\sum_{d\leq x} \sum_{n \leq x, d|n} \frac{1}{d}= \sum_{d\leq x} \frac{1}{d} \sum_{n \leq x, d|n} 1 =\sum_{d\leq x} \frac{1}{d} \lfloor \frac {x}{d}\rfloor $ but we see that d is a dummy variable so it follows that $\sum_{n\leq x} \frac{\sigma (n)}{n} = \sum_{n \leq x} \frac{1}{n} \lfloor \frac{x}{n} \rfloor $

From here we have that $\sum_{n \leq x} \frac{1}{n} \lfloor \frac{x}{n} \rfloor = \sum_{n \leq x} \frac{x}{n^2} + O(1) = x \sum_{n =1}^{\infty} \frac{1}{n^2} + \sum_{n \leq x} O(1) - \sum_{x}^{\infty} \frac {1}{x} = x \space \zeta(2) + \sum_{n \leq x} O(1) - O(\log (x))$ It follows that $x \frac{\pi^2}{6} + \sum_{n \leq x} O(1) + O(\log (x))$ but something wrongs as it appears that $\sum_{n \leq x} O(1) $ is $O(x)$ which is not what i want.

Answer 1

You can get an explicit bound on the error term.

Since $\frac{x}{n}-1 \le \lfloor \frac{x}{n} \rfloor \le \frac{x}{n}$,

$\sum_{n \leq x} \frac{1}{n} \lfloor \frac{x}{n} \rfloor \le \sum_{n \leq x} \frac{1}{n} \frac{x}{n} = x\sum_{n \leq x} \frac{1}{n^2} $

and

$\sum_{n \leq x} \frac{1}{n} \lfloor \frac{x}{n} \rfloor \ge \sum_{n \leq x} \frac{1}{n} (\frac{x}{n}-1) = x\sum_{n \leq x} \frac{1}{n^2} -\sum_{n \leq x} \frac{1}{n} = x\sum_{n \leq x} \frac{1}{n^2} -\ln(x)+O(1) $

Answer 2

I think you're just getting your notation a bit mixed up at the end. You say $$\sum_{n \leq x} \frac{1}{n} \left\lfloor \frac{x}{n} \right\rfloor = \sum_{n \leq x} \frac{x}{n^2} + O(1)$$ I think you want the $O(1)$ under the summation sign:

$$\sum_{n \leq x} \frac{1}{n} \left\lfloor \frac{x}{n} \right\rfloor = \sum_{n \leq x} \frac{1}{n}\left(\frac{x}{n} + O(1)\right)$$

and from there on, it looks like plain sailing.

Answer 3

You have

$$\sum_{n \leq x} \frac{\sigma(n)}{n} = \ldots = \sum_{n \leq x} \frac{1}{n} \left\lfloor\frac{x}{n}\right\rfloor = \sum_{n \leq x} \frac{x}{n^2} + O(1)$$

but this should be

$$\sum_{n \leq x} \frac{1}{n} \left\lfloor\frac{x}{n}\right\rfloor = \sum_{n \leq x} \frac{1}{n} \left[ \frac{x}{n} + O(1) \right] = x \sum_{n \leq x} \frac{1}{n^2} + O(1) \sum_{n \leq x} \frac{1}{n}$$

You can probably take it from there.