TITLE: How to prove the inequality $\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n-1}\geq \log (2)$?
QUESTION [10 upvotes]: Prove the following inequality:
$$\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n-1}\geq \log (2)$$
I have tried experimenting with different values of $n$ and see that the sum seems to converge to $\log(2)$ as $n$ gets larger, but I am having some difficulty proving this inequality. I realize it is probably something to do with the fact that $\frac{d}{dx}\ln(x)=\frac{1}{x}$, but cannot find a proper solution. 
The theme of these problems is that they can generally be solved with some sort of drawing or visual aid, and I am unsure of what I can draw to make this solution more clear.
Any help is appreciated, thanks.

REPLY [17 votes]: It's well known that the following inequality holds:
$$e^x\geqslant x+1\geqslant 0,\ \forall\ x\geqslant 0.$$
So we have $$\displaystyle\prod_{i=n}^{2n-1}e^{{1\over i}}\geqslant \prod_{i=n}^{2n-1}\big(1+{1\over i}\big)=\prod_{i=n}^{2n-1}{i+1\over i}={2n\over n}=2.$$
Thus $$\displaystyle\sum_{i=n}^{2n-1}{1\over i}\geqslant \ln2.$$