Prove that a series is bounded with induction I have to prove that the following condition is true:
$$\frac{1}{n+1} + \frac{1}{n+2} + ... + \frac{1}{2n} > \frac{13}{24}$$
for every $n > 1$.
I understood that this series is the same as:
$$S(n) = \sum_{i=1}^n \frac{1}{n + i} $$
I tried to use induction but I could't reach something meaningful. 
(This exercise was in a discrete math exam, so I am not looking for a proof using series theory).
Thanks!
 A: Let
$$
S(n) = \frac{1}{n+1} + \frac{1}{n+2} + \ldots + \frac{1}{2n}.
$$
So,
$$
S(n+1) = \frac{1}{n+2} + \frac{1}{n+3} + \ldots + \frac{1}{2n} + \frac{1}{2n+1} + \frac{1}{2n+2}
$$
and
$$
S(n+1) = S(n) - \frac{1}{n+1} + \frac{1}{2n+1} + \frac{1}{2n+2} = S(n) + \frac{1}{2n+1} - \frac{1}{2n+2} =\\= S(n) + \frac{1}{(2n+1)(2n+2)}
$$
For $n=2$ we have
$$
S(2) = \frac13 + \frac14 = \frac{7}{12} = \frac{14}{24} > \frac{13}{24},
$$
and $S(n)$ strictly increasing. So...?
A: Hint:
$$\frac12=1-\frac12,\quad\frac13+\frac14=1-\frac12+\frac13-\frac14,\ldots$$
What can you conclude from this?
A: By Cauchy-Schwarz inequality, 
$\displaystyle\left(\sum_{i=1}^n \dfrac{1}{n+i}\right) \cdot \displaystyle \left(\sum_{i=1}^n n+i \right) \geq n^2$  
$\implies S(n) \geq \dfrac{n^2}{n^2+\frac{n(n+1)}{2}} $  
$\implies S(n) \geq \dfrac{2n^2}{3n^2+n}$  
Since $f(n)=\left(\dfrac{2n^2}{3n^2+n} -\dfrac{13}{24}\right)$ is a strictly increasing function for $n>1$ ,  
$\displaystyle \boxed {\therefore {S(n)>\frac{13}{24} \ \forall \ n>1}}$
