Let $\sigma$ the application that transforms $n$ into the sum of its divisors (ex : $\sigma\left(6\right)=12$)\ I've proved that $$ n+1 \leq \sigma\left(n\right) \leq n+n\ln\left(n\right) $$ I know that $$ \frac{\sum_{k=1}^{n}\sigma\left(k\right)}{n^2} \underset{n \rightarrow +\infty}{\rightarrow}\frac{\pi^2}{12} $$ I just want to show that this sequence converges (because I guess the limit value $\pi^2/12$ is not that easy to neatly get with elementary math). I think that this sequence decreases, and it would be sufficient for me to show it but I havent been successful in proving it.
Any hint ?
It is not obvious to me that $\frac{\sum_{k\le n}\sigma(k)}{n^2}$ may be decreasing. It is immediate that it converges from
$$\sum_{k\le n}\sigma(k)=\sum_{k\le n}\sum_{dm=k}d=\sum_{dm\le n} d=\sum_{m\le n} \frac{\lfloor n/m\rfloor (\lfloor n/m\rfloor+1)}{2}= n^2 \sum_{m\le n} \frac{1}{2m^2}+O(\sum_{m\le n} n/m)$$