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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.

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  • 1
    $\begingroup$ You can write \sum_{k=1}^{2n-1}\frac1k\geq\log(2) to generate $$\sum_{k=1}^{2n-1}\frac1k\geq\log(2)$$ or for a more “modern” inequality sign, you can write \geqslant to generate $\geqslant$ $\endgroup$ – Mr Pie Jan 15 '18 at 4:24
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Use Riemann sum. Observe \begin{align} \frac{1}{n} + \ldots + \frac{1}{2n-1}\geq \int^{2n}_{n} \frac{1}{x}\ dx = \log 2n-\log n = \log 2. \end{align}

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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 $$\displaystyle\sum_{i=n}^{2n-1}{1\over i}\geqslant \ln2.$$

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