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

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 $n=k+1$:

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

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

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 $7\cdot24>12\cdot13$

Your sum ($=:S_n$) can be written as follows: $$S_n=-{1\over2n}+\left({1\over 2n}+{1\over n+1}+{1\over n+2}+\ldots+{1\over 4n}\right)+{1\over 4n}\ .$$ Now the large parenthesis can be viewed as a trapezoidal sum for the integral $\int_n^{2n}{1\over x}\ dx=\log 2$. Since the function $x\mapsto{1\over x}$ is convex for $x>0$ the trapezoidal sum is greater than the integral; therefore we immediately obtain $$S_n>\log 2-{1\over 4n}\geq\log2-{1\over8}\doteq0.568\qquad(n\geq2)\ .$$ Since ${13\over24}\doteq0.542$ the claim follows.