Question: I've been trying to evaluate properly the following limit
$$\lim\_{n\to \infty}\frac{H\_n}{n}$$
where $H\_n$ is the $n$-th harmonic number $\left(H\_n=\sum\_{i=1}^n\frac1i\right)$.

My guess is that the answer is $0$, but I would appreciate rigorous explanation.

Answer: 
> 
> With [Stolz–Cesàro Theorem](http://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem) :
> 
> 
> 

$$
\lim\_{n\ \to\ \infty}{H\_{n + 1} - H\_{n} \over \left(\,n + 1\,\right) - n}
=\lim\_{n\ \to\ \infty}{1 \over n + 1} = 0\qquad\Longrightarrow\qquad\color{#66f}{\large\lim\_{n\ \to\ \infty}{H\_{n} \over n}} = \color{#66f}{\Large 0}
$$
**Hint**

For large values of $n$, the harmonic number can be approximated by $$H\_n=\gamma +\log (n)+O\left(\left(\frac{1}{n}\right)\right)$$ So, taking into account the respective behavior of $\log(n)$ and $n$, the limit you are looking for is $0$. 

This is an expansion to remember since very often used in the context of similar questions.