Prove that: $\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n}\ge\frac{2}{3}$ I've got three inequalities:
$\forall n\in\mathbb N:$
$$\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n} \ge\frac{1}{2}$$
$$\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n} \ge\frac{7}{12}$$
$$\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n}\ge\frac{2}{3}$$
From what I know the LHS converges to something about $0.69$ and each one of them requires the same method, but I can't come up with a proper way to solve it.

Can someone give me a hint?

 A: Hint: $$\frac{1}{n}\ge\ln(n+1)-\ln(n)$$
A: We know that $$\color{red}{ \frac{x}{x+1}\le\ln(x+1)\le x~~~\forall ~x>0\tag{1}\label{eq}}$$
taking $x=\frac{1}{n+k}~~0\le k\le n$ this lead to $$\ln\left(\frac{1}{n+k}+1\right)\le\frac{1}{n+k}~~~\forall ~~0\le k\le n $$
that is 
\begin{split}
\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n} &=&\sum_{k=0}^{n} \frac{1}{n+k} \ge  \sum_{k=0}^{n} \ln\left(\frac{1}{n+k}+1\right)\\&=&\sum_{k=0}^{n} \ln\left(n+k+1\right)-\ln\left(n+k\right)\\ &=&\ln\left(2n+1\right)-\ln\left(n\right)  \\&= &\ln\left( \frac{2n+1}{n}\right) =  \ln\left( 1 +\frac{n+1}{n}\right) \end{split}
Then, Using the left side of $\eqref{eq}$ 
$$\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n}  \ge  \ln\left( 1 +\frac{n+1}{n}\right)\ge\frac{n+1}{n+2} \ge  \frac{2}{3} $$
Since for all $n\ge 1$ $$\frac{n+1}{n+2} \ge \frac{1+1}{1+2}  = \frac{2}{3}$$

Because $x\mapsto \frac{x+1}{x+2}$ is an incresing function with derivative $\frac{1}{(x+2)^2}>0. $

Finally observes that
 $$\frac{2}{3}>\frac{7}{12}>\frac{1}{2}$$
