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Question Number 108371 by Skabetix last updated on 16/Aug/20

$$provebyreccurence$$ $$\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2n}<or=\frac{n}{2}$$ $$thanks$$

Answered by Aziztisffola last updated on 16/Aug/20

$$\mathrm{for}\mathrm{n}=1\Rightarrow \frac{1}{2}⩽\frac{1}{2}\mathrm{true}$$ $$\mathrm{suppose}\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2n}⩽\frac{\mathrm{n}}{2}$$ $$\mathrm{and}\mathrm{prove}\mathrm{that}\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2n}+\frac{1}{2\mathrm{n}+2}⩽\frac{\mathrm{n}+1}{2}$$ $$\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2n}+\frac{1}{2\mathrm{n}+2}⩽\frac{\mathrm{n}}{2}+\frac{1}{2\mathrm{n}+2}$$ $$\frac{\mathrm{n}}{2}+\frac{1}{2\mathrm{n}+2}=\frac{\mathrm{n}}{2}+\frac{1}{2(\mathrm{n}+1)}⩽\frac{\mathrm{n}}{2}+\frac{1}{2}=\frac{\mathrm{n}+1}{2}$$ $$\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2n}+\frac{1}{2\mathrm{n}+2}⩽\frac{\mathrm{n}+1}{2}$$ $$\mathrm{hence}\mathrm{\forall }\mathrm{n}\in {\mathbb{N}}^{\ast }\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\frac{1}{2\mathrm{k}}⩽\frac{\mathrm{n}}{2}$$

Answered by 1549442205PVT last updated on 16/Aug/20

$$\mathrm{i})\mathrm{For}\mathrm{n}=1\mathrm{we}\mathrm{have}\frac{1}{2}=\frac{1}{2}\Rightarrow \mathrm{true}$$ $$\mathrm{ii})\mathrm{Consider}\mathrm{n}⩾5\mathrm{we}\mathrm{have}$$ $$\frac{1}{10}<\frac{1}{4.5}=\frac{1}{4}-\frac{1}{5},\frac{1}{12}<\frac{1}{5.6}=\frac{1}{5}-\frac{1}{6}...$$ $$\frac{1}{2\mathrm{n}}<\frac{1}{(\mathrm{n}-1)\mathrm{n}}=\frac{1}{\mathrm{n}-1}-\frac{1}{\mathrm{n}}$$ $$\Rightarrow \frac{1}{10}+\frac{1}{12}+...+\frac{1}{2\mathrm{n}}<\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...$$ $$...+\frac{1}{\mathrm{n}-1}-\frac{1}{\mathrm{n}}=\frac{1}{4}-\frac{1}{\mathrm{n}}.\mathrm{Hence},$$ $$\mathrm{Adding}\mathrm{up}\mathrm{we}\mathrm{get}$$ $$\mathrm{LHS}<(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8})+\frac{1}{4}-\frac{1}{\mathrm{n}}<\frac{\mathrm{n}}{2}$$ $$\iff \frac{12+6+4+3}{24}-\frac{1}{\mathrm{n}}<\frac{\mathrm{n}}{2}\iff \frac{25}{24}-\frac{1}{\mathrm{n}}<\frac{\mathrm{n}}{2}$$ $$\iff \frac{25\mathrm{n}-24}{24\mathrm{n}}<\frac{\mathrm{n}}{2}\iff {12\mathrm{n}}^2-25\mathrm{n}+24>0$$ $$\iff 12{(\mathrm{n}-\frac{25}{24})}^2+24-\frac{625}{48}>0$$ $$\mathrm{is}\mathrm{true}\mathrm{\forall }\mathrm{n}⩾2$$ $$\mathrm{Consequently},$$ $$\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2\mathit{n}}⩽\frac{\mathbf{n}}{2}(\mathbf{q}.\mathbf{e}.\mathbf{d})$$ $$$$

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