Proving that $\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n}>\frac{13}{24}$ by induction. Where am I going wrong? I have to prove that
$$\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{2n}>\frac{13}{24}$$ for every positive integer $n$.
After I check the special cases $n=1,2$, I have to prove that the given inequality holds for the $n+1$ case by using the $n$ case. So, I have to prove:
$$\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{2n+2}\overset{?}{>}\frac{13}{24}$$
what is equivavlent to 
$$\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{2n+2}\overset{?}{>}\frac{13}{24}+\frac{1}{n}$$
By using the $n$ case, I know that 
$$\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{2n+2}>\frac{13}{24}+\frac{1}{2n+1}+\frac{1}{2n+2}$$
so I basically have to prove that
$$\frac{13}{24}+\frac{1}{2n+1}+\frac{1}{2n+2}\overset{?}{>}\frac{13}{24}+\frac{1}{n}$$
what is equivavlent to 
$$\frac{1}{2n+1}+\frac{1}{2n+2}\overset{?}{>}\frac{1}{n}$$
$$n(2n+2)+n(2n+1)\overset{?}{>}(2n+1)(2n+2)$$
$$4n^2+3n\overset{?}{>}4n^2+6n+2$$
$$-3n\overset{?}{>}2$$
$$n\overset{?}{<}-\frac{2}{3}$$
which can not be because $n$ is a positive integer. What am I doing wrong? Tnx!

As I see it, here's what might be confusing you:
If you let
$$A=\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{2n+2},$$
and
$$B=\frac{13}{24}+\frac{1}{n},$$
then you are trying to prove that $A>B$. Also, if you let
$$C=\frac{13}{24}+\frac{1}{2n+1}+\frac{1}{2n+2},$$
then you have by assumption that $A>C$. Thus if you can prove that $C>B$, as you write yourself, then your are finished, since this will imply that $A>B$.
The problem with this however is that $C\not > B$. This is why you get a contradiction, when you try to prove it, and there's nothing wrong with this. We can have both $A>B$ and $A>C$ without $C>B$.
