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Question Number 117895 by aurpeyz last updated on 14/Oct/20

$$provebymathematicalinductionthat$$ $$\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}>\frac{1}{2}$$

Commented byDwaipayan Shikari last updated on 14/Oct/20

$$\frac{n+1+n+2+...}{n}⩾\frac{n}{\frac{1}{n+1}+\frac{1}{n+2}+....}$$ $$\frac{2n^2+n^2+n}{2n^2}⩾\frac{1}{\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...}\left(WithoutMathematicalInduction\right)$$ $$\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...⩾\frac{2n^2}{3n^2+n}$$ $$\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...⩾\frac{2n}{3n+1}$$ $$GeneralInequality$$ $$Ifwetaken=1$$ $$then$$ $$\frac{1}{1+1}+\frac{1}{1+2}+\frac{1}{1+3}+....>\frac{2}{4}is\left(Alwaysgreaterthan\frac{1}{2}\right)$$ $$So$$ $$\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+....>\frac{1}{2}$$

Answered by 1549442205PVT last updated on 14/Oct/20

$$\mathrm{Prove}\mathrm{that}\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}>\frac{1}{2}\left(1\right)\mathrm{for}\mathrm{n}>2,\mathrm{n}\in \mathrm{N}$$ $$\mathrm{i})\mathrm{For}\mathrm{n}=2\mathrm{we}\mathrm{have}\frac{1}{2+1}+\frac{1}{2+2}=\frac{1}{3}+\frac{1}{4}$$ $$=\frac{7}{12}>\frac{6}{12}=\frac{1}{2}\Rightarrow \mathrm{The}\mathrm{inequality}\mathrm{is}\mathrm{true}$$ $$\mathrm{ii})\mathrm{Suppose}\mathrm{the}\mathrm{inequality}\mathrm{was}\mathrm{true}$$ $$\mathrm{for}\mathrm{n}=\mathrm{k}\mathrm{that}\mathrm{means}{\mathrm{S}}_{\mathrm{k}}\underset{\mathrm{p}=1}{=\mathrm{\Sigma }}\frac{1}{\mathrm{k}+\mathrm{p}}<\frac{1}{2\mathrm{k}}$$ $$\mathrm{iii})\mathrm{Need}\mathrm{prove}\mathrm{that}{\mathrm{S}}_{\mathrm{k}+1}=\underset{\mathrm{p}=1}{\stackrel{\mathrm{k}}{\mathrm{\Sigma }}}\frac{1}{\mathrm{k}+1+\mathrm{p}}<\frac{1}{2(\mathrm{k}+1)}$$ $$\mathrm{Indeed},{\mathrm{S}}_{\mathrm{k}+1}=\frac{1}{\mathrm{k}+2}+...+\frac{1}{2\mathrm{k}}+\frac{1}{2(\mathrm{k}+1)}$$ $$=\frac{1}{\mathrm{k}+1}+\frac{1}{\mathrm{k}+2}+...+\frac{1}{2\mathrm{k}}+(\frac{1}{2(\mathrm{k}+1)}-\frac{1}{\mathrm{k}+1})$$ $$={\mathrm{S}}_{\mathrm{k}}-\frac{1}{2(\mathrm{k}+1)}<\frac{1}{2\mathrm{k}}-\frac{1}{2(\mathrm{k}+1)}=\frac{\mathrm{k}+1-\mathrm{k}}{2\mathrm{k}(\mathrm{k}+1)}$$ $$=\frac{1}{2\mathrm{k}(\mathrm{k}+1)}<\frac{1}{2(\mathrm{k}+1)}.\mathrm{This}\mathrm{shows}\mathrm{the}$$ $$\mathrm{inequality}\left(1\right)\mathrm{is}\mathrm{also}\mathrm{true}\mathrm{for}\mathrm{n}=\mathrm{k}+1$$ $$\mathrm{By}\mathrm{induction}\mathrm{mathematic}\mathrm{principle}$$ $$\mathrm{it}\mathrm{is}\mathrm{true}\mathrm{\forall }\mathrm{n}\in \mathrm{N},\mathrm{n}⩾2$$ $$\mathbf{second}\mathbf{way}\left({\mathrm{don}}^{\prime }\mathrm{t}\mathrm{use}\mathrm{induction}\mathrm{method}\right)$$ $$\mathrm{Since}\mathrm{n}+\mathrm{k}<2\mathrm{n}\mathrm{\forall }\mathrm{k}=1,2,...,(\mathrm{n}-1),\mathrm{so}$$ $$\frac{1}{\mathrm{n}+\mathrm{k}}<\frac{1}{2\mathrm{n}}\mathrm{\forall }\mathrm{k}=1,2,...,(\mathrm{n}-1).\mathrm{Hence}$$ $$\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}>\frac{1}{2\mathrm{n}}+\frac{1}{2\mathrm{n}}+...+\frac{1}{2\mathrm{n}}$$ $$=\frac{1+1+...+1}{2\mathrm{n}}=\frac{\mathrm{n}}{2\mathrm{n}}=\frac{1}{2}$$ $$\mathrm{Since}\mathrm{the}\mathrm{sum}\mathrm{consist}\mathrm{of}\mathrm{n}\mathrm{terms}$$

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