Solve the following

Let $\mathrm{p}(n): \frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2 n}>\frac{13}{24}$, for all natural numbers $n>1$.

Step I: For $n=2$,

$\mathrm{LHS}=\frac{1}{2+1}+\frac{1}{2 \times 2}=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}=\frac{14}{24}>\frac{13}{24}=\mathrm{RHS}$

As, LHS $>$ RHS

So, it is true for $n=2$.

Step II : For $n=k$,

Let $\mathrm{p}(k): \frac{1}{k+1}+\frac{1}{k+2}+\ldots+\frac{1}{2 k}>\frac{13}{24}$, be true for some natural numbers $k>1$.

Step III : For $n=k+1$,

$\mathrm{p}(k+1)=\frac{1}{k+1}+\frac{1}{k+2}+\ldots+\frac{1}{2 k}+\frac{1}{2 k+1}$

$>\frac{13}{24}+\frac{1}{2 k+1} \quad($ Using step II $)$

$>\frac{13}{24}$

i.e. $\mathrm{p}(k+1)>\frac{13}{24}$

So, it is also true for $n=k+1$.

Hence, $\mathrm{p}(n): \frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2 n}>\frac{13}{24}$, for all natural numbers $n>1$.