1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Did you now that $\log(n+1)\leq H_n\leq 1+\log n$? $\endgroup$ – String Sep 15 '14 at 16:12
  • $\begingroup$ I'm not familiar with the inequality. Can you provide some reference. $\endgroup$ – Scippy Sep 15 '14 at 16:17
2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

With Stolz–Cesàro Theorem :

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.