This question has been raised because of my initial confusion concerning this question on the upper Darboux sum of $f$:
Again consider $f : [0,1] \mapsto \mathbb R$ be a function defined as $$ f(x) = \begin{cases} x & \text{if x is rational} \\ 0 & \text{if x is irrational} \end{cases} $$
However in this case I don't want to consider this function as continuous and integrate it, I want to consider it a discrete function consisting of an infinite number of discrete rational fractions and create a suitable weighted mean sum over the interval $[0,1]$.
I wish to calculate the limit of the following weighted mean sum as $n$ tends to infinity:
$$S_n=\frac{1}{n}\sum _{m=2}^n \frac{1}{m-1}\left(\sum _{k=1}^{m-1} \frac{k}{m}-\frac{\sigma _1(m)-m-1}{m}\right)\tag{1}$$
where $\sigma _1(m)$ is the sum of all the divisors of $m$ and $(\sigma _1(m)-m-1)$ is equivalent to Chowla's function for $m>1$
Now using the approximation $$S_n <\frac{1}{n}\sum _{m=2}^n \frac{1}{m-1}\sum _{k=1}^{m-1} \frac{k}{m}=\frac{1}{2}-\frac{1}{2 n}$$
in the limit $n \to \infty$
$$S_{n \to \infty} < \frac{1}{2}$$
However is it seems that in actual fact for my sum $(1)$
$$S_{n \to \infty}=\frac{1}{2}$$
which would mean
$$\lim_{n \to \infty} \frac{1}{n}\sum _{m=2}^n \left(-\frac{\sigma _1(m)-m-1}{(m-1)m}\right)=0 \,.\tag{2}$$
However I'm not sure how to prove this last statement and the relation of the weighted sum formula (1) to the notional Riemann Integral over rational $f$ assuming continuity?
Presumably I need to first obtain an upper bound somehow for Chowla's function for $m>1$
The discrete plot of
$$\frac{1}{n}\sum _{m=2}^n \left(\frac{\sigma _1(m)-m-1}{(m-1)m}\right) \tag{3}$$
is shown below:
Compared with a discete plot below using a bounding function from Jeffrey Lagarias assuming the Riemann Hypothisis is true: $$\frac{1}{n}\sum _{m=2}^n \frac{H_m+e^{H_m} \log \left(H_m\right)-m-1}{m (m-1)}$$
I'm not sure the bound needs to be this good to prove $(2)$, even though it seems a better bound to (3) might conjecturally be
$$\frac{1}{e\, n}\sum _{m=2}^n \frac{H_m+e^{H_m} \log \left(H_m\right)-m-1}{m (m-1)}$$
As a further note; numerically it appears that
$$\lim_{n \to \infty} \frac{1}{n}\sum _{m=1}^n \frac{\sigma _1(m)}{m}\approx \frac{\pi^2}{6}$$
which may help with finding a bounding function for $\sigma _1(m)$.
The conjecture of course is that:
$$\lim_{n \to \infty} \frac{1}{n}\sum _{m=1}^n \frac{\sigma _1(m)}{m}= \frac{\pi^2}{6}$$
Update 2:
I can get a very good approximation to
$$\frac{1}{n}\sum _{m=2}^n \frac{\sigma _1(m)}{m (m-1)}$$
using for example
$$\frac{\pi ^2}{6 n} \sum _{m=2}^n \frac{\left(m-\frac{1}{10}\right)^2}{m^2(m-1)}$$
where $\frac{1}{10}$ is a constant added to help give a bounding curve fit. However how do I justify making use of this (admittedly in a sightly modified form) or similar approximation, to prove the limit in $(2)$.