Two inequalities for the difference $H_n - H_m$ of harmonic numbers

Question

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?

Answer 1

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?

Answer 2

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 $\int_1^{\infty} \frac{dx}{x}$