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  • √n<1/√1+1/√2+ ....+ 1/√n for all natural numbers n ≥ 2.

\sqrt{n}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots .+\frac{1}{\sqrt{n}}  for all natural numbers n \geq 2.

Answers (1)

P(n) is \sqrt{n}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots .+\frac{1}{\sqrt{n}}  ; n \geq 2.

\mathrm{P}(2) is \sqrt{2}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}=1.414<1.7 .0 \ldots thus, it is true\\ \mathrm{P}(3)\ is \sqrt{3}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}=1.732<2.284,\ thus \ it is\ true\\ Now, let us consider P(k)=\sqrt{k}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots .+\frac{1}{\sqrt{k}},\ is \ true\\ Now, we will add \sqrt{k+1}\ on\ both \ sides,

Thus,

\begin{array}{l} \sqrt{k}+\sqrt{k+1}-\sqrt{k}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots .+\frac{1}{\sqrt{k}}+\sqrt{k+1}-\sqrt{k} \\ \text { Thus, } \sqrt{k+1}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots .+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}} \end{array}

Hence, by mathematical induction,

For each natural no. n \geq 2,P(n) is \sqrt{n}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots .+\frac{1}{\sqrt{n}}

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infoexpert21

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