TITLE: Proof of $\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\frac{\sigma(k)}{k}=\zeta(2)$
QUESTION [6 upvotes]: Numerically, they seem to be the same, but can we prove that
$$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\frac{\sigma(k)}{k}=\zeta(2),$$
where $\sigma(k)$ is the sum of divisors of $k$.

REPLY [11 votes]: Notice that $\frac{\sigma(k)}{k} = \displaystyle \sum_{d | k} \frac{1}{d}$, since $\sigma(k)$ is just the sum of the divisors of $k$.  
Then we can write
$$ \frac{1}{n} \sum_{k=1}^n \frac{\sigma(k)}{k}
 = \frac{1}{n} \sum_{k=1}^n \left(\sum_{d | k} \frac{1}{d}\right) 
 = \frac{1}{n} \sum_{d=1}^n \frac{\left[\frac{n}{d}\right]}{d} $$
since the term $\frac{1}{d}$ shows up once for each $k$ from 1 to $n$ that is a multiple of $d$.
Taking the limit as $n \rightarrow \infty$, we get
$$ \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{d=1}^n \frac{\left[\frac{n}{d}\right]}{d} = \sum_{d=1}^\infty \frac{1}{d^2} 
= \zeta(2). $$