Question

Prove that $\frac{1}{n+1}+\frac{1}{n+2}+....+\frac{1}{2n}>\frac{13}{24}$, for all natural numbers n > 1.

Answer 1

$\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}>\frac{13}{24}$,

Using induction we first show this is true for n = 2 : $\frac{1}{3}+\frac{1}{4}=\frac{7}{12}=\frac{14}{24}>\frac{13}{24}\left(True\right)$

Now lets assume it is true for some n = k,

$S_k=\frac{1}{k+1}+\frac{1}{k+2}+....+\frac{1}{2k}>\frac{13}{24}$

Finally we need to prove that this implies it is also true for n = k + 1 ;

$s_{k+1}=\frac{1}{k+2}+\frac{1}{k+3}+....+\frac{1}{2k+2}$

$=\frac{-1}{k+1}+\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3}+.....+\frac{1}{2k}+\frac{1}{2k+1}+\frac{1}{2k+2}$

$=\frac{-1}{k+1}+S_k+\frac{1}{2k+1}+\frac{1}{2k+2}$

$=S_k+\frac{1}{2(2k+1)(k+1)}>S_k$

$\therefore S_{k+1}>\frac{13}{24}$