Let $F(n)=\sum\limits_{d\mid n} \frac{1}{d}\;$ then prove that $\lim_{N\to\infty} \frac{1}{N}\sum\limits_{n=1}^N F(n)=\frac{\pi^2}{6}$. Let $F(n)=\sum\limits_{d\mid n} \frac{1}{d}\;$  then prove that $\lim_{N\to\infty} \frac{1}{N}\sum\limits_{n=1}^N  F(n)=\frac{\pi^2}{6}$.
i know the result of this .is we processed using this result
$\lim\limits_{n\to\infty}\left[\frac{1}{n}\sum_{i=1}^{n}f\left(\frac{i}{n}\right)\right]=\int_{0}^{1}f(x)\,{\rm d}x$
 A: Just an explanation for comments by Shanchul Lee and user1952009. We consider the sum
$$\sum_{n=1}^N F(n)=\sum_{n=1}^N\sum_{d\mid n}\frac 1 d$$
Fix some $d$. How many times does $1/d$ appear in the summation? It's not hard to see that this number is equal to the number of $n$ divisible by $d$ where $n\le N$, which can be expressed as $\lfloor N/d\rfloor$. So we may rewrite the sum as
$$\sum_{d=1}^N \bigg\lfloor\frac N d\bigg\rfloor\frac 1 d$$ 
Now note that $|\lfloor N/d\rfloor-N/d|\leq1$, so
\begin{align}
&\left|\frac 1 N \sum_{n=1}^N F(n) - \sum_{d=1}^N \frac 1 {d^2}\right|\\
=&\frac 1 N \left|\sum_{d=1}^N \frac 1 d \left(\bigg\lfloor\frac N d\bigg\rfloor-\frac N d\right)\right|\\
\leq& \frac 1 N\sum_{d=1}^N \frac 1 d=\mathcal O (\log N/N).
\end{align}
Here we invoked the well-known fact that $\sum_{k=1}^N 1/k=\mathcal O(\log n)$.
Now let $N\to\infty$, since $\lim\limits_{N\to\infty}\log N/N=0$, we get
$$\lim_{N\to\infty}\frac 1 N \sum_{n=1}^N F(n)=\sum_{d=1}^\infty \frac 1 {d^2}=\frac{\pi^2} 6$$
A: For future reference observe that
$$F(n) = \frac{1}{n}\sum_{d|n} \frac{n}{d}
= \frac{1}{n} \sigma(n).$$
Now we have
$$\sum_{n\ge 1} \frac{\sigma(n)}{n^s} = \zeta(s)\zeta(s-1).$$
Hence
$$\sum_{n\ge 1} \frac{F(n)}{n^s} = \zeta(s+1)\zeta(s).$$
Hence               by               the               Wiener-Ikehara
theorem
we have
$$\frac{1}{N} \sum_{n\le N} F(n)
\sim \frac{1}{N} \frac{N^1}{1} \mathrm{Res}_{s=1} \zeta(s+1)\zeta(s)
= \frac{\pi^2}{6}.$$
