Question

For any natural number n > 1
$\frac{1}{n+1}+\frac{1}{n+2}+.....+\frac{1}{2n}>\frac{13}{14}$

Answer 1

Let $S_n=\frac{1}{n+1}+\frac{1}{n+2}+....+\frac{1}{2}$
For $n=2$, we have
$\frac{1}{3}+\frac{1}{4}=\frac{7}{12}=\frac{14}{24}>\frac{13}{24}$
Hence the inequality holds for $n=2$
Now assume $S_m>\frac{13}{24}$ for some positive integer $m>1$
We have $S_m=\frac{1}{m+1}+\frac{1}{m+2}+....+\frac{1}{2m}$
and $S_{m+1}=\frac{1}{m+2}+\frac{1}{m+3}+....+\frac{1}{2m}+\frac{1}{2m+1}+\frac{1}{2m+2}$
$\therefore S_{m+1}-S_m=\frac{1}{2m+1}+\frac{1}{2m+2}-\frac{1}{m+1}$
that is , $\therefore S_{m+1}-S_m=\frac{1}{2(m+1)(2m+1)}>0$ for any natural number m . It follows that $\therefore S_{m+1}-S_m>0,thatis{S}_{m+1}>S_m$
But $S_m>\frac{13}{24}$ by our assumption
$\therefore S_{m+1}>\frac{13}{24}$
Hence by mathematical induction . The given inequality holds for all natural numbers n >1