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  • Prove that 1/(n+1)+ 1/(n+2)+ .....+ 1/2n> 13/24,for all natural numbers n > 1.

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

Answers (1)

Given:

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

Now, we’ll substitute different values for n,

\begin{aligned} \text { At } n=2 & \\ P(2)=1 / 2 &+1 / 4 \\ &=3 / 4>13 / 24 \end{aligned} \\ It\: \: is\: \: true.\\ At \: \: $n=3$ \\ \begin{aligned} P(3)=1 / 2 &+1 / 4+1 / 6 \\ &=11 / 12>13 / 24 \end{aligned}\\.

It is true.
Now, let us consider,

\\P(k)=\frac{1}{k+1}+\frac{1}{k+2}+\ldots \ldots+\frac{1}{2 k}>\frac{13}{24}$ To be true.\\ Now, at $n=k+1$ ,\\$P(k+1)=\frac{1}{(k+1)+1}+\frac{1}{(k+1)+2}+\ldots . .+\frac{1}{2(k+1)}

\begin{array}{l} =\frac{1}{k+2}+\frac{1}{k+3} \ldots \ldots+\frac{1}{2 k+2} \\\\ =\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3} \ldots \ldots+\frac{1}{2 k}+\frac{1}{2 k+1}+\frac{1}{2 k+2}-\frac{1}{k+1} \\\\ =\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3} \ldots \ldots+\frac{1}{2 k}+\frac{2k+2+2k+1-4k-2}{2(2 k+1)(k+1)} \\\\ =\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3} \ldots \ldots+\frac{1}{2 k}+\frac{1}{2(2 k+1)(k+1)}>\frac{13}{24} \end{array}

Thus, n = k + 1 is true

Thus, by mathematical Induction,

For each natural no. n > 1 it is true that,

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

Posted by

infoexpert21

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