Proving $\frac{1}{n+1} + \frac{1}{n+2}+\cdots+\frac{1}{2n} > \frac{13}{24}$ for $n>1,n\in\Bbb N$ by Induction

Proving $\frac{1}{n+1} + \frac{1}{n+2}+\cdots+\frac{1}{2n} > \frac{13}{24}$ for $n>1,n\in\Bbb N$

To solve it I used induction but it is leading me nowhere my attempt was as follows:

Lets assume the inequality is true for $n = k$ then we need to prove that it is true for $k+1$

so we need to prove $\frac1{k+2} + \frac1{k+3}+\cdots+\frac1{2(k+1)} > 13/24$

I don't know where to go from here please help.

Solution 1:

Using induction we first show this is true for $n=2$:

$\frac{1}{2+1}+\frac{1}{2+2}=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}=\frac{14}{24}\gt\frac{13}{24}$

Therefore it is indeed true for $n=2$.

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

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

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

$S_{k+1}=\frac{1}{(k+1)+1}+\frac{1}{(k+1)+2}+...+\frac{1}{2(k+1)-2}+\frac{1}{2(k+1)-1}+\frac{1}{2(k+1)}$
$\qquad=\frac{1}{k+2}+\frac{1}{k+3}+...+\frac{1}{2k}+\frac{1}{2k+1}+\frac{1}{2(k+1)}$
$\qquad=-\frac{1}{k+1}+\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3}+...+\frac{1}{2k}+\frac{1}{2k+1}+\frac{1}{2(k+1)}$
$\qquad=-\frac{1}{k+1}+S_k+\frac{1}{2k+1}+\frac{1}{2(k+1)}$
$\qquad=S_k+\frac{1}{2k+1}+\frac{1}{2(k+1)}-\frac{1}{k+1}$
$\qquad=S_k+\frac{1}{2(2k+1)(k+1)}$
$\qquad\gt S_k$

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

Solution 2:

Let $$f(n)=\frac1{n+1}+\frac1{n+2}+\cdots+\frac1{2n}=\sum_{r=1}^{2n}\frac1r-\sum_{s=1}^n\frac1s$$

So, $$f(n+1)-f(n)=\frac1{(2n+1)(2n+2)}>0$$ for integer $n\ge0$

$\displaystyle \implies f(n)$ is an increasing function.

Now, $f(2)=\frac13+\frac14=\frac7{12}>\frac{13}{24}$ as $7\cdot24>12\cdot13$