Use the Principle of Mathematical Induction to prove that $\frac{1}{n+1} + \frac{1}{n+2} + \ldots + \frac{1}{2n} > \frac{13}{24}$ for all natural numbers $n > 1$.

(A) Let $P(n): \frac{1}{n+1} + \frac{1}{n+2} + \ldots + \frac{1}{2n} > \frac{13}{24}$.
Step $1$: For $n = 2$,the expression is $\frac{1}{2+1} + \frac{1}{2+2} = \frac{1}{3} + \frac{1}{4} = \frac{7}{12} = \frac{14}{24}$.
Since $\frac{14}{24} > \frac{13}{24}$,$P(2)$ is true.
Step $2$: Assume $P(k)$ is true for some $k \in N, k > 1$,i.e.,$\frac{1}{k+1} + \frac{1}{k+2} + \ldots + \frac{1}{2k} > \frac{13}{24}$.
Step $3$: For $n = k+1$,we need to show $P(k+1) > \frac{13}{24}$.
$P(k+1) = \frac{1}{k+2} + \frac{1}{k+3} + \ldots + \frac{1}{2k} + \frac{1}{2k+1} + \frac{1}{2k+2}$.
We can write $P(k+1) = P(k) - \frac{1}{k+1} + \frac{1}{2k+1} + \frac{1}{2k+2}$.
$P(k+1) = P(k) + \frac{1}{2k+1} + \frac{1}{2(k+1)} - \frac{1}{k+1} = P(k) + \frac{1}{2k+1} - \frac{1}{2(k+1)}$.
Since $2k+2 > 2k+1$,it follows that $\frac{1}{2k+1} > \frac{1}{2k+2}$,so $\frac{1}{2k+1} - \frac{1}{2(k+1)} > 0$.
Thus,$P(k+1) > P(k) > \frac{13}{24}$.
Therefore,$P(k+1)$ is true whenever $P(k)$ is true.
By the Principle of Mathematical Induction,$P(n)$ is true for all $n > 1$.