Prove that $1/n+1/(n+1)+\dots+(1/2n)>2/3$ 
Prove that $\displaystyle\frac{1}{n} + \frac{1}{n+1} + \dots + \frac{1}{2n}>\frac{2}{3}$

I tried to use mathematical induction, but I'm not able to prove that:
$$
\frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{2n} + \frac{1}{2n+1} + \frac{1}{2n+2}>\frac{2}{3}.
$$
My method was:
Assumption:
$$\frac{1}{n} + \frac{1}{n+1} + \dots + \frac{1}{2n}>\frac{2}{3}$$
$$\frac{1}{n} + \frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{2n} + \frac{1}{2n+1} + \frac{1}{2n+2}>\frac{2}{3} + \frac{1}{2n+1} + \frac{1}{2n+2}$$
$$\frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{2n} + \frac{1}{2n+1} + \frac{1}{2n+2}>\frac{2}{3} + \frac{1}{2n+1} + \frac{1}{2n+2}-\frac{1}{n}$$
Now in order to prove the thesis I have to prove that
$$\frac{2}{3} + \frac{1}{2n+1} + \frac{1}{2n+2}-\frac{1}{n} > \frac{2}{3}$$
But it's a contradiction. Did I make a mistake somewhere? How can I solve this problem? I'd appreciate your help.
 A: HINT
As suggested by Wojowu in the comment, sometimes induction works for a stronger hypothesis, in that case let try with
$$\displaystyle\frac{1}{n} + \frac{1}{n+1} + \dots + \frac{1}{2n}>\frac{2}{3}+\frac1{4n}>\frac23$$
Refer also to the related


*

*Proof by induction, $1/2 + ... + n/2^n < 2$

*If $x>0$ real number and $n>1$ integer, then $(1+x)^n>\frac{1}{2}n(n-1)x^2$

*Show that $\left(1+\frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\left(1+\frac{1}{3^3}\right)...\left(1+\frac{1}{n^3}\right) < 3$
A: There is also a "non-inductive" way to show the inequality.
You may transform the sum into a Riemann-sum and then estimate this sum by an integral:
$$1/n+1/(n+1)+\dots+(1/2n) = \frac{1}{n}\left(\frac{1}{1+\frac{0}{n}} + \frac{1}{1+\frac{1}{n}} + \cdots + \frac{1}{1+\frac{n}{n}} \right) =\boxed{\sum_{k=0}^n \frac{1}{1+\frac{k}{n}} \cdot \frac{1}{n}}$$
Now, note that $f(x) = \frac{1}{1+x}$ is strictly decreasing $[0,1]$. So, you have


*

*$\frac{1}{1+\frac{k}{n}} \cdot \frac{1}{n} = \int_{\frac{k}{n}}^{\frac{k+1}{n}} \frac{1}{1+\frac{k}{n}} \; dx > \int_{\frac{k}{n}}^{\frac{k+1}{n}} \frac{1}{1+x}\; dx$
This gives
$$\boxed{\sum_{k=0}^n \frac{1}{1+\frac{k}{n}} \cdot \frac{1}{n} > \int_0^1 \frac{1}{1+x}\; dx = \ln 2 > \frac{2}{3}}$$
