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Prove that \(\frac{1}{n+1}+\frac{1}{n+2}+\).........+ \(\frac{1}{2n}>\frac{13}{24}\),for all natural numbers n > 1.

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Let,P(n) =  \(\frac{1}{n+1}+\frac{1}{n+2}+\).........+ \(\frac{1}{2n}>\frac{13}{24}\) ∀ natural numbers n > 1.

Step1: For n = 2

\(\frac{1}{2+1}+\frac{1}{2+2}\) = \(\frac{1}3\)\(\frac{1}4\) = \(\frac{1}{2n}>\frac{13}{24}\)

So, it is true for n = 2 

Step2: For n = k

P(k) = \(\frac{1}{k+1}+\frac{1}{k+2}\)+......+ \(\frac{1}{2k}>\frac{13}{24}\)

Now, we need to show that P(k+1) is true when P(k) is true.

P(k) = \(\frac{1}{k+2}+\frac{1}{k+3}\)+......\(\frac{1}{2k}+\frac{1}{2(k+1)}\)

As, LHS = RHS 

Thus, P(k+1) is true. So, by the principle of mathematical induction 

P(n) is true for all n.

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