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Question Number 113070 by ZiYangLee last updated on 11/Sep/20

$$\mathrm{If}n\in {\mathbb{Z}}^{+},\mathrm{prove}\mathrm{that}$$ $$\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...\frac{1}{(n+1)\sqrt{n}}<2$$ $$$$

Answered by 1549442205PVT last updated on 11/Sep/20

$$\mathrm{We}\mathrm{have}\frac{1}{(\mathrm{n}+1)\sqrt{\mathrm{n}}}=\sqrt{\mathrm{n}}.\frac{1}{(\mathrm{n}+1)\mathrm{n}}$$ $$=\sqrt{\mathrm{n}}(\frac{1}{\mathrm{n}}-\frac{1}{\mathrm{n}+1})=\sqrt{\mathrm{n}}(\frac{1}{\sqrt{\mathrm{n}}}+\frac{1}{\sqrt{\mathrm{n}+1}})(\frac{1}{\sqrt{\mathrm{n}}}-\frac{1}{\sqrt{\mathrm{n}+1}})$$ $$=(1+\frac{1}{\sqrt{\mathrm{n}+1}})(\frac{1}{\sqrt{\mathrm{n}}}-\frac{1}{\sqrt{\mathrm{n}+1}})$$ $$\mathrm{Since}(1+\frac{1}{\sqrt{\mathrm{n}+1}})<2,\mathrm{so}\left[\frac{1}{(\mathrm{n}+1)\sqrt{\mathrm{n}}}\right]$$ $$<2(\frac{1}{\sqrt{\mathrm{n}}}-\frac{1}{\sqrt{\mathrm{n}+1}}).\mathrm{Hence},\mathrm{give}\mathrm{n}=1,2,...$$ $$\mathrm{we}\mathrm{get}$$ $$\frac{1}{2\sqrt{1}}=\frac{1}{2}<2(\frac{1}{1}-\frac{1}{\sqrt{2}}),\frac{1}{3\sqrt{2}}<2(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}),$$ $$...,\frac{1}{(\mathrm{n}+1)\sqrt{\mathrm{n}}}<2(\frac{1}{\sqrt{\mathrm{n}}}-\frac{1}{\sqrt{\mathrm{n}+1}})$$ $$\mathrm{Adding}\mathrm{up}\mathrm{n}\mathrm{above}\mathrm{inequalities}\mathrm{we}\mathrm{get}$$ $$\mathrm{LHS}=\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...\frac{1}{(n+1)\sqrt{n}}<$$ $$2[\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{\mathrm{n}}}-\frac{1}{\sqrt{\mathrm{n}+1}})$$ $$=2(1-\frac{1}{\sqrt{\mathrm{n}+1}})<2$$ $$\mathbf{Therefore},\mathbf{we}\mathbf{have}$$ $$\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...\frac{1}{(\mathit{n}+1)\sqrt{\mathit{n}}}<2(\mathbf{q}.\mathbf{e}.\mathbf{d})$$

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