# 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.

• Did you now that $\log(n+1)\leq H_n\leq 1+\log n$? – String Sep 15 '14 at 16:12
• I'm not familiar with the inequality. Can you provide some reference. – Scippy Sep 15 '14 at 16:17

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