I have recently learned about asymptotical equivalence, defined as $$ a(n) \sim b(n) \Leftrightarrow \lim\limits_{n \to \infty} \frac{a(n)}{b(n)} = 1$$

Now I would need to prove that $H_n \sim \ln(n)$ where $H_n = \sum_{i=1}^n \frac{1}{i}$ is the harmonic series and $\ln(n)$ is the natural logarithm, but I have no idea how to get started.

I wrote down the definition as \begin{align} \lim\limits_{n \to \infty} \frac{H_n}{\ln(n)} & = \lim\limits_{n \to \infty} \left(\frac{1}{\ln(n)} \cdot \sum_{i=1}^n \frac{1}{i}\right) \\ \end{align} and tried to apply l'Hospital's rule which leads to $\lim\limits_{n \to \infty} \left(\frac{-1}{n\ln^2(n)} \cdot \sum_{i=1}^n \frac{-1}{i^2}\right)$, but that does not seem to be helping me any further. I tried out some other things, but everything seemed to be incorrect or not helpful.

I am looking at it for quite some time now already and I also tried to find some hints online, but I just don't seem to find anything that uses this definition. I must admit I am not really the biggest hero in analysis or calculus, but if anyone could guide me in (one of) the right directions to get to a solution that would already be very helpful. Thanks in advance.

  • 1
    $\begingroup$ L'Hopital's rule won't work here because a sum is a discrete sequence and therefore has no derivative. However, there is a "discrete L'Hopital's rule" which works here. That's what Jack uses in his answer below. $\endgroup$ – Antonio Vargas Apr 9 '16 at 15:54

By the Stolz-Cesàro theorem, $$ \lim_{n\to +\infty}\frac{H_n}{\log n} = \lim_{n\to +\infty}\frac{H_{n+1}-H_{n}}{\log(n+1)-\log n} =\lim_{n\to +\infty}\frac{1}{(n+1)\log\left(1+\frac{1}{n}\right)}=\color{red}{1}.$$

As an alternative, $$ 0\leq H_n-\log(n+1) = \sum_{k=1}^{n}\left(\frac{1}{k}-\log\left(1+\frac{1}{k}\right)\right)\leq\sum_{k=1}^{n}\frac{1}{2k^2}<\frac{\zeta(2)}{2}=\frac{\pi^2}{12} $$ is enough. It comes from a telescoping trick and the fact that $x-\log(1+x)\leq\frac{x^2}{2}$ over $[0,1]$.

  • $\begingroup$ That's really a useful theorem. I find it strange that I've never heard about it before... I wanted to try to understand the alternative, but I'm afraid I'm missing a bit too much in order to get it, but thanks anyway! If I get some spare time I have something new to think about ;) $\endgroup$ – Mr Tsjolder Apr 8 '16 at 19:07

It is

$H_n - 1 = \sum_{k=2}^{n} \frac{1}{k} \leq \sum_{k=2}^{n} \int_{k-1}^{k} \frac{1}{k} = \int_{1}^{n} \frac{1}{x} \mathrm{d}x = \ln (x)$


$H_n - \frac{1}{n} = \sum_{k=1}^{n-1} \frac{1}{k} \geq \sum_{k=1}^{n-1} \int_{k}^{k+1} \frac{1}{x} = \ln(x)$.

Hence $\ln (n) + \frac{1}{n} \leq H_n \leq \ln (n) + 1$.

Therefore $1 + \frac{1}{n \ln (n)} \leq \frac{H_n}{\ln (n)} \leq 1 + \frac{1}{\ln (n)}$.

You will get the desired result by $n \to \infty$.


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