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!

Let
$$ A_n = \frac{1}{n}+\frac{1}{n+1}+\ldots+\frac{1}{2n}=H_{2n}-H_{n-1}.\tag{1}$$
Then:
$$ A_n-A_{n+1} = \frac{1}{n}-\frac{1}{2n+1}-\frac{1}{2n+2}>0 \tag{2} $$
hence the sequence $\{A_n\}$ is decreasing and to prove our claim it is enough to show that:
$$ \lim_{n\to +\infty}A_n >\frac{13}{24}=0.541666\ldots.\tag{3}$$
However, by a Riemann sum argument:
$$ \lim_{n\to +\infty}A_n = \lim_{n\to +\infty}\frac{1}{n}\left(\frac{1}{1}+\frac{1}{1+\frac{1}{n}}+\frac{1}{1+\frac{2}{n}}+\ldots+\frac{1}{2}\right)\\=\int_{1}^{2}\frac{dx}{x}=\log 2=0.69314718\ldots\tag{4}$$
and we are done. We may also notice that from the upper bound:
$$ A_n-A_{n-1} = \frac{2+3n}{2n(n+1)(2n+1)}\leq\frac{3}{(2n)(2n+1)}\tag{5}$$
it follows that:
$$ A_n \geq \frac{3}{2}-\sum_{n\geq 1}\frac{3}{(2n)(2n+1)}=\log 8-\frac{3}{2}=0.579441\ldots\tag{6}$$
without using any integral.
