<s>[INST] How prove this $H_{2n}-H_{n}+\frac{1}{4n}>\ln{2}$ 
Show that, for every positive integer $n$,
  $$\dfrac{1}{n+1}+\dfrac{1}{n+2}+\cdots+\dfrac{1}{2n}+\dfrac{1}{4n}>\ln{2}$$

I know this
$$\lim_{n\to\infty}\dfrac{1}{n+1}+\dfrac{1}{n+2}+\cdots+\dfrac{1}{2n}=\ln2$$
and use this 
$$\ln{(1+\dfrac{1}{n})}<\dfrac{1}{n}$$ is not useful
But for this inequality I can't. Thank you
 [/INST] Write $$u_n = \frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{2n} + \frac{1}{4n}$$
$$ = [\frac{1}{1} + \frac{1}{2} + \dots+ \frac{1}{4n}] - [\frac{1}{1} + \frac{1}{2} + \dots+ \frac{1}{n}] - [\frac{1}{2n+1} + \frac{1}{2n+2} + \dots +\frac{1}{4n -1}]$$
$$= [\gamma_{4n} + \log(4n)] - [\gamma_n + \log(n)] - [\gamma_{4n -1} + \log(4n-1) - \gamma_{2n} - \log(2n)]$$ 
Now take $\lim_{n \rightarrow \infty} u_n$.
All $\gamma_{n}$, $\gamma_{4n - 1}$, $\gamma_{2n}$, $\gamma_{n}$ etc will be $\gamma$ (Euler's constant) and will be cancelled out.
What remains?
$\lim_{n \rightarrow \infty}[\log{\frac{4n}{4n - 1}} + \log{\frac{2n}{n}}] = \log 2$
Now see $u_{n+1} - u_n = \frac{1}{4(n+1)} - \frac{1}{4n} + \frac{1}{2(n+1)} - \frac{1}{n+1} < 0$
Thus the sequence $\{u_n\}$ is monotone decreasing and converges to $\log 2$. So for all $n$, $u_n > \log 2$
Thus ultimately we are getting for any $n$
$$\frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{2n} + \frac{1}{4n} > \log 2$$
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