2
$\begingroup$

Let's note the harmonic numbers by $H_n = \sum_{i=1}^n \frac{1}{i}$.

Asymptotic expansion of harmonic number is: $(1) H_n = \ln n + \gamma + \frac{1}{2n} - O\left(\frac{1}{n^2}\right)$

Very popular is the inequality: $(2)\ln n < H_n < \ln n + 1$

Assuming $m < n$, using $(1)$ we get: $$H_n - H_{m} = \ln(n) - \ln(m) + \frac{1}{2n} - \frac{1}{2m} < \ln(n) - \ln(m)$$

I try to use $(2)$ similarly and get:

$$H_n - H_m < \ln(n) + 1 - H_m < \ln(n) + 1 - \ln(m)$$

Can someone give me a hint,why I have two different results?

$\endgroup$
  • 2
    $\begingroup$ You want the sum to start at $i=1$, not $i=0$. $\endgroup$ – Robert Israel Apr 29 '14 at 15:21
  • $\begingroup$ @RobertIsrael Yes, of course you are right. Thanks! $\endgroup$ – Rop Apr 29 '14 at 15:36
3
$\begingroup$

Since $0 < \gamma < 1$, that is certainly true.

EDIT: For the new question, I don't see what your problem is. It's certainly true that for $m < n$, $$ H_n - H_m < \ln(n) - \ln(m) + \dfrac{1}{n} - \dfrac{1}{m} + O(1/n^2) + O(1/m^2)$$ It's also true that for $m < n$, $$H_n - H_m = \sum_{k=m+1}^n \dfrac{1}{k} < \sum_{k=m+1}^n \int_{k-1}^k \dfrac{dt}{t} = \int_{m}^n \dfrac{dt}{t} = \ln(n) - \ln(m)$$ And this implies the weaker inequality $$H_n - H_m < \ln(n) + 1 - \ln(m)$$ So what?

$\endgroup$
0
$\begingroup$

I think it's not an approximation but exact inequality. That is, it holds without any $O(...)$.

To see that, you can bound the series with an integral $\int_1^{\infty} \frac{dx}{x}$

$\endgroup$

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.