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: 
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.
**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.