How to show that $\lim \frac{1}{n} \sum_{i=1}^n \frac{1}{i}=0 $?

$$\sum_{i=1}^n \frac{1}{i} = \sum_{1\le i\le\sqrt{n}} \frac{1}{i} + \sum_{\sqrt{n}<i\le n} \frac{1}{i} \le \sum_{1\le i\le\sqrt{n}} 1 + \sum_{\sqrt{n}<i\le n} \frac{1}{\sqrt{n}} \le \sqrt{n} + \sqrt{n} = 2\sqrt{n}.$$

We can approximate a finite sum with a definite integral (see here). We obtain that $$ \log(n+1)=\int_1^{n+1}x^{-1}\mathrm dx\le\sum_{i=1}^n\frac1i\le1+\int_1^nx^{-1}\mathrm dx=1+\log n. $$ Now we need to show that $$ \lim_{n\to\infty}\frac{\log n}n=0. $$ This can be done by using l'Hôpital's rule (a more general statement is proved here).

In the same direction as user121270 and coolydudey60 in their comments $$ \frac{1}{n} \sum_{i=1}^n \frac{1}{i} =\frac{H_n}{n}$$ and for large values of $n$ $$\frac{H_n}{n}=\frac{\gamma +\log \left(n\right)}{n}+\frac{1}{2 n^2}-\frac{1}{12 n^3}+O\left(\left(\frac{1}{n}\right)^4\right)$$ and then what V.C. proposed in his answer $\frac{1+\log(n)}{n}$ seems to be very good.