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Solution
\[\text{ Step I: For n } = 2, \]
\[LHS = \frac{1}{2 + 1} + \frac{1}{2 \times 2} = \frac{1}{3} + \frac{1}{4} = \frac{7}{12} = \frac{14}{24} > \frac{13}{24} = RHS\]
\[As, LHS > RHS\]
\[\text{ So, it is true for n } = 2 . \]
\[\text{ Step II: For n } = k, \]
\[\text{ Let } p\left( k \right): \frac{1}{k + 1} + \frac{1}{k + 2} + . . . + \frac{1}{2k} > \frac{13}{24}, \text{ be true for some natural numbers } k > 1 . \]
\[\text{ Step III: For n } = k + 1, \]
\[p\left( k + 1 \right) = \frac{1}{k + 1} + \frac{1}{k + 2} + . . . + \frac{1}{2k} + \frac{1}{2k + 1}\]
\[ > \frac{13}{24} + \frac{1}{2k + 1} \left( \text{ Using step } II \right)\]
\[ > \frac{13}{24}\]
\[i . e . p\left( k + 1 \right) > \frac{13}{24}\]
\[\text{ So, it is also true for n } = k + 1 . \]
\[\text{ Hence,} p\left( n \right): \frac{1}{n + 1} + \frac{1}{n + 2} + . . . + \frac{1}{2n} > \frac{13}{24}, \text{ for all natural numbers} n > 1 .\]
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