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
 A: Let $A_n=H_{2n}-H_{n}+\frac{1}{4n}$, then a tedious but easy computation shows that
$$
A_{n+1}-A_n=\frac{-1}{4n(n+1)(2n+1)}\lt0,
$$
hence $(A_n)$ is decreasing, in particular, $A_n\gt A$ where $A=\lim\limits_{n\to\infty}A_n$. Since $H_n=\log n+\gamma+o(1)$ when $n\to\infty$, $A=\log2$. QED.
Remark: The funny thing here (and the subject of the next questions of the exercise as it is usually asked) is that the same technique applies to the sequence $B_n=H_{2n}-H_{n}$, showing that $(B_n)$ is increasing, hence $B_n\lt B$ where $B=\lim\limits_{n\to\infty}B_n$. Since $B=\log2$ as well, this yields some approximations of $\log2$, namely the fact that, for every $n\geqslant1$,
$$
H_{2n}-H_{n}\lt\log2\lt H_{2n}-H_{n}+\frac1{4n}.
$$
Likewise, for every $n\geqslant1$,
$$
H_{2n}-H_{n}+\frac1{4n+2}\lt\log2\lt H_{2n}-H_{n}+\frac1{4n},
$$
thus, the correction $\frac1{4n}$ is asymptotically of the right order.
A: A not at all tedious computation shows that the expression - which I rearrange a tiny bit -
$$\begin{align}
\frac{1}{2n} + \frac{1}{n+1} + \dotsc + \frac{1}{2n-1} + \frac{1}{4n} &=  \sum_{k=0}^{n-1} \frac{1}{2}\left(\frac{1}{n+k} + \frac{1}{n+k+1}\right)\\
&= \frac{1}{n} \sum_{k=0}^{n-1} \frac{1}{2}\left(\frac{1}{1 + \frac{k}{n}} + \frac{1}{1+\frac{k+1}{n}}\right)
\end{align}$$
is a trapezium sum for the integral
$$\int_0^1 \frac{dt}{1+t} = \log 2,$$
and since $\frac{1}{1+t}$ is convex, the trapeziums have greater area than the corresponding part of the integral.
A: 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 $n$
$$\frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{2n} + \frac{1}{4n} > \log 2$$
