TITLE: Showing $\sum_{k=1}^{nm} \frac{1}{k} \approx \sum_{k=1}^{n} \frac{1}{k} + \sum_{k=1}^{m} \frac{1}{k}$
QUESTION [13 upvotes]: Since $\log(nm) = \log(n) + \log(m)$, and $\sum_{k=1}^n \frac{1}{k} \approx \log n$ for large $n$, we would expect that $$\sum_{k=1}^{nm} \frac{1}{k} \approx \sum_{k=1}^{n} \frac{1}{k} + \sum_{k=1}^{m} \frac{1}{k}$$
when $n,m$ are large. 
I'm wondering if this approximation can be demonstrated through discrete means. That is to say, manipulations of rational fractions and/or elementary number-theoretical considerations, without using the $\log n$ approximation.

REPLY [2 votes]: Avoiding $\log$s we still have $H_N=\sum_{k=1}^{N}\frac{1}{k}=\int_{0}^{1}\frac{x^N-1}{x-1}$ and 
$$ H_{NM}-H_{N}-H_{M} = \int_{0}^{1}\frac{(x^N-1)(x^M-1)}{x-1}\,dx\leq 0, $$
whose absolute value can be bounded (due to the Cauchy-Schwarz inequality) by
$$ \sqrt{\int_{0}^{1}\frac{(1-x^N)^2}{1-x}\,dx \int_{0}^{1}\frac{(1-x^M)^2}{1-x}\,dx}=\sqrt{(2H_N-H_{2N})(2H_M-H_{2M})}. $$
On the other hand
$$ H_{2N}-H_N = \sum_{k=1}^{2N}\frac{(-1)^{k+1}}{k} $$
is a partial sum for a convergent (to $\log 2$) series, hence
$$ H_N+H_M-\sqrt{H_N H_M}\leq H_{NM} \leq H_N+H_M  $$
holds for any large $N,M$.