How to prove this inequality?$\frac{1}{n+1} + \frac{1}{n+2} + \cdots+\frac{1}{n+n} + \frac{1}{4n} > \ln 2$

Question

$$\frac{1}{n+1} + \frac{1}{n+2} + \cdots +\frac{1}{n+n} + \frac{1}{4n} > \ln 2$$

$n$ is positive integer.

Thank you !

Answer 1

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:

$$ \frac{2}{\frac{1}{4n+4}+\frac{1}{4n}} < \frac{4n+4n+4}{2}=4n+2$$

Answer 2

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.