﻿ If n\geq 1 Proof of the square root inequality 2\sqrt{n+1}-2\sqrt{n}<\dfrac{1}{\sqrt{n}}<2\sqrt{n}-2\sqrt{n-1}

#### Question

If n\geq 1 Proof of the square root inequality

2\sqrt{n+1}-2\sqrt{n}<\dfrac{1}{\sqrt{n}}<2\sqrt{n}-2\sqrt{n-1}

Collected in the board: Inequality

Steven Zheng posted 2 years ago

Since f(x) = \sqrt{x} is an increasing function,

\sqrt{n+1} > \sqrt{n}>\sqrt{n-1}

Add \sqrt{n} in each side, which will not change the direction of the inequality sign

\sqrt{n+1} + \sqrt{n} > 2\sqrt{n}>\sqrt{n-1}+ \sqrt{n}

(\sqrt{n+1} + \sqrt{n})\cdotp \dfrac{\sqrt{n+1} - \sqrt{n}}{\sqrt{n+1} - \sqrt{n}} > 2\sqrt{n}>(\sqrt{n-1}+ \sqrt{n})\cdotp \dfrac{\sqrt{n-1}- \sqrt{n}}{\sqrt{n-1}- \sqrt{n}}

\dfrac{1}{\sqrt{n+1} - \sqrt{n}} > 2\sqrt{n}>\dfrac{1}{\sqrt{n}-\sqrt{n-1} }

Taking the reciprocal in the inequality leads to change of the inequality sign.

\sqrt{n+1} - \sqrt{n} <\dfrac{1}{2\sqrt{n} }<\sqrt{n}-\sqrt{n-1}

Therefore,

2\sqrt{n+1} -2 \sqrt{n} <\dfrac{1}{\sqrt{n} }<2\sqrt{n}-2\sqrt{n-1}

Steven Zheng posted 2 years ago

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