Induction inequality on sum of reciprocals I have to prove that: $\displaystyle\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}\ge\frac{1}{2}$ for natural $n$ 
Checking for $n=1$ we have $\displaystyle 1+\frac{1}{2}=\frac{3}{2}\ge \frac{1}{2}$ 
Next we assume that inequality is true for $n$ and for $n+1$ we have: 
$\displaystyle\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}+\frac{1}{2n+1}\ge\frac{1}{2}+\frac{1}{2n+1}\ge\frac{1}{2}$ what is true because $\displaystyle \frac{1}{2n+1} \ge 0$
Is my proof correct ? 

The proof is nearly correct, but induction is unnecesary:
$$\frac{1}{n} + \frac{1}{n+1} +\ldots + \frac{1}{2n} \geq \underbrace{\frac{1}{2n} + \frac{1}{2n}+\ldots +\frac{1}{2n} \frac{1}{2n}}_{n\text{ times}} = n\frac{1}{2n} = \frac{1}{2}$$
