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 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.
