<s>[INST] Proof for $\log\frac{2n+1}{n+1}<\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}<\log 2$ How can I prove that
$\displaystyle \log\frac{2n+1}{n+1}<\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}<\log2$
using $\displaystyle \frac{x}{1+x}<\log(1+x)<x$
 [/INST] Riemann sums do this pretty quickly. $\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}$ is a 
lower Riemann sum for $\int_n^{2n} {1 \over x}\,dx = \log 2$, and an upper Riemann sum for 
$\int_{n+1}^{2n + 1} {1 \over x}\,dx = \log {2n +1 \over n + 1}$, so one has
$$\log {2n +1 \over n + 1} < \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} < \log 2 $$
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