How to prove this inequality?$\frac{1}{n+1} + \frac{1}{n+2} + \cdots+\frac{1}{n+n} + \frac{1}{4n} > \ln 2$ $$\frac{1}{n+1} + \frac{1}{n+2} + \cdots +\frac{1}{n+n} + \frac{1}{4n} > \ln 2$$
$n$ is positive integer.
Thank you !
 A: Because $\frac 1t$ is concave up, its integral is overestimated by the trapezoidal approximation.  In this light, we have
$$
\begin{align}
\ln 2 = \int_n^{2n}\frac 1t dt &<
\frac 12 \left( \frac 1{n} + \frac 1{n+1} \right) +
\frac 12 \left( \frac 1{n+1} + \frac 1{n+2} \right) + \cdots+ 
\frac 12 \left( \frac 1{2n-1} + \frac 1{2n} \right)\\
&= \frac 1{2n} + \frac 12 \left( \frac 1{n+1} + \frac 1{n+1} \right) + 
\frac 12 \left( \frac 1{n+2} + \frac 1{n+2} \right) + \cdots \\
&\qquad+
\frac 12 \left( \frac 1{2n-1} + \frac 1{2n-1} \right) + \frac 12 \frac 1{2n}
\\
&= \frac{1}{n+1} + \frac{1}{n+2} + ... +\frac{1}{2n} + \frac{1}{4n} &=
\end{align}
$$
I hope that's clear.
A: Let 
$$a_n=\frac{1}{n+1} + \frac{1}{n+2} + \cdots +\frac{1}{n+n} + \frac{1}{4n}$$
Then 
$$a_{n+1}-a_n=\frac{1}{2n+1}+\frac{1}{2n+2}+\frac{1}{4n+4}-\frac{1}{n+1}-\frac{1}{4n}\\
 =\frac{1}{2n+1}+\frac{2}{4n+4}+\frac{1}{4n+4}-\frac{4}{4n+4}-\frac{1}{4n}\\
 =\frac{2}{4n+2}-\frac{1}{4n+4}-\frac{1}{4n} <0 $$
Thus, $a_n$ is a strictly decreasing sequence. As $\lim_n a_n = \ln 2$, it follows that 
$$a_n > \ln(2)$$
P.S.
$$\frac{2}{4n+2} < \frac{1}{4n+4}+\frac{1}{4n}$$
can be easily proven by bringing everything to the same denominator, but it also follows from the HM-AM inequality:
$$ \frac{2}{\frac{1}{4n+4}+\frac{1}{4n}} < \frac{4n+4n+4}{2}=4n+2$$
