Limit of the difference between two harmonic numbers

Question

I am trying to evaluate the following limit:

$$ \lim_{n \rightarrow \infty} H_{kn} - H_n = \lim_{n \rightarrow \infty} \sum_{i = 1}^{(k - 1)n} \frac{1}{n + 1} $$

I know that the answer is $\ln k$, but I have no idea how to even approach this problem. I have tried somehow equating this to the Taylor expansion of $\ln(1 + (k - 1))$, but to no avail.

Here is the case $k=2$: The limit of truncated sums of harmonic series, $\lim\limits_{k\to\infty}\sum_{n=k+1}^{2k}{\frac{1}{n}}$ — Martin Sleziak, Feb 13 '18 at 00:29

score 2 · Accepted Answer · answered Feb 13 '18 at 00:32

2

HINT:

Use the Riemann sum to evaluate

$$H_{kn}-H_n =\sum_{m=1}^{(k-1)n}\frac1{n+m}=\frac1n \sum_{m=1}^{(k-1)n}\frac{1}{1+m/n}$$

answered Feb 13 '18 at 00:32

Mark Viola

179,405

score 1 · Answer 2 · answered Feb 13 '18 at 00:35

1

Hint Use the fact that $$\lim_{n} H_{kn}-\ln(kn)=\lim_n H_n-\ln(n)=\gamma$$

answered Feb 13 '18 at 00:35

N. S.

132,525

score 1 · Answer 3 · answered Feb 13 '18 at 05:23

1

Use the asymptotics $$H_p=\gamma +\log \left({p}\right)+\frac{1}{2 p}+O\left(\frac{1}{p^2}\right)$$ making $$H_{kn}-H_n=\log \left({k}\right)+\frac{\frac{1}{k}-1}{2 n}+O\left(\frac{1}{n^2}\right)$$

answered Feb 13 '18 at 05:23

Claude Leibovici

260,315

Limit of the difference between two harmonic numbers

3 Answers3