Limit involving harmonic number 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.
 A: Note that by the Riemann definition of integral (noting the positiveness of the functions involved), by taking  $\Delta x = 1$ and calculating the upperbound of the integral in each interval, you could see that $1 + \int\limits_2^{n + 1} {\frac{{dx}}{{x - 1}}}  \ge \sum\limits_{i = 1}^n {\frac{1}{i}} $, that is $$\sum\limits_{i = 1}^n {\frac{1}{i}}  \le 1 + \ln (n)$$Now you should be able to proceed to the desired result.
A: 
With Stolz–Cesàro Theorem :

$$
\lim_{n\ \to\ \infty}{H_{n + 1} - H_{n} \over \left(\,n + 1\,\right) - n}
=\lim_{n\ \to\ \infty}{1 \over n + 1} = 0\qquad\Longrightarrow\qquad\color{#66f}{\large\lim_{n\ \to\ \infty}{H_{n} \over n}} = \color{#66f}{\Large 0}
$$
A: 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.
