Limiting value of $\frac{\sigma(n)}{n}$

Question

I am having some trouble with the following:

Let $\sigma(n)$ be the sum of the positive divisors of $n$, e.g. $\sigma(6)=1+2+3+6=12$. What is the 'expected value' of the abundancy index $\frac{\sigma(n)}{n}$? I.e., if you pick an integer 'at random', and compute this abundancy index, what limiting average value will we get?

With regard to 'picking an integer at random', the spirit of the book suggests an interpretation along the lines of calculating $\lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^n \frac{\sigma(k)}{k}.$

One possible approach-

$\sigma$ is a multiplicative function, i.e. $\sigma(ab)=\sigma(a)\sigma(b)$ for $a$, $b$ relatively prime.

Multiplicative functions $f$ satisfy

$\sum_{k=1}^\infty \dfrac{f(k)}{k^s}=\prod_{p \text{ prime}}\left(1+\dfrac{f(p)}{p^s} + \dfrac{f(p^{2})}{p^{2s}} + \ ...\right)$

This is convenient here since we have a nice expression for primes, $\sigma(p^k)=1+p+p^2+ \ ...\ p^k$.

I am not sure if this will give a proof.

Answer 1

Hint: For a start,

$$ \sum_{k=1}^n \frac{\sigma(k)}{k} = \sum_{k=1}^n \sum_{d|k} \frac{d}{k}$$ Now switch the order of summation.

Answer 2

$$\frac{1}{n}\sum_{k \le n} \frac{\sigma(k)}{k} = \frac{1}{n}\sum_{k \le n} \frac{1}{k}\sum_{d \mid k} d = \frac{1}{n}\sum_{d \le n} d \sum_{d \mid k \le n} \frac{1}{k} = \frac{1}{n}\sum_{d \le n} d\sum_{j \le n/d} \frac{1}{dj}$$ $$ = \frac{1}{n}\sum_{d \le n} [\log(n/d)+O(1)] = \log(n)-\frac{1}{n}\sum_{d \le n} \log(d)+O(1) = O(1).$$

So I guess you need to keep track of constants, which is not something I want to do now.