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Question Number 117895 by aurpeyz last updated on 14/Oct/20
prove by mathematical induction that (1/(n+1))+(1/(n+2))+...+(1/(2n))>(1/2)
$${prove}\:{by}\:{mathematical}\:{induction}\:{that} \\ $$$$\frac{\mathrm{1}}{{n}+\mathrm{1}}+\frac{\mathrm{1}}{{n}+\mathrm{2}}+…+\frac{\mathrm{1}}{\mathrm{2}{n}}>\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by Dwaipayan Shikari last updated on 14/Oct/20
((n+1+n+2+...)/n)≥(n/((1/(n+1))+(1/(n+2))+....)) ((2n^2 +n^2 +n)/(2n^2 ))≥(1/((1/(n+1))+(1/(n+2))+(1/(n+3))+...)) (Without Mathematical Induction) (1/(n+1))+(1/(n+2))+(1/(n+3))+...≥((2n^2 )/(3n^2 +n)) (1/(n+1))+(1/(n+2))+(1/(n+3))+...≥((2n)/(3n+1)) General Inequality If we take n=1 then (1/(1+1))+(1/(1+2))+(1/(1+3))+....>(2/4) is (Always greater than (1/2)) So (1/(n+1))+(1/(n+2))+(1/(n+3))+....>(1/2)
$$\frac{{n}+\mathrm{1}+{n}+\mathrm{2}+…}{{n}}\geqslant\frac{{n}}{\frac{\mathrm{1}}{{n}+\mathrm{1}}+\frac{\mathrm{1}}{{n}+\mathrm{2}}+….} \\ $$$$\frac{\mathrm{2}{n}^{\mathrm{2}} +{n}^{\mathrm{2}} +{n}}{\mathrm{2}{n}^{\mathrm{2}} }\geqslant\frac{\mathrm{1}}{\frac{\mathrm{1}}{{n}+\mathrm{1}}+\frac{\mathrm{1}}{{n}+\mathrm{2}}+\frac{\mathrm{1}}{{n}+\mathrm{3}}+…}\:\:\:\left({Without}\:{Mathematical}\:{Induction}\right) \\ $$$$\frac{\mathrm{1}}{{n}+\mathrm{1}}+\frac{\mathrm{1}}{{n}+\mathrm{2}}+\frac{\mathrm{1}}{{n}+\mathrm{3}}+…\geqslant\frac{\mathrm{2}{n}^{\mathrm{2}} }{\mathrm{3}{n}^{\mathrm{2}} +{n}} \\ $$$$\frac{\mathrm{1}}{{n}+\mathrm{1}}+\frac{\mathrm{1}}{{n}+\mathrm{2}}+\frac{\mathrm{1}}{{n}+\mathrm{3}}+…\geqslant\frac{\mathrm{2}{n}}{\mathrm{3}{n}+\mathrm{1}} \\ $$$${General}\:{Inequality} \\ $$$${If}\:{we}\:{take}\:{n}=\mathrm{1} \\ $$$${then} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{3}}+….>\frac{\mathrm{2}}{\mathrm{4}}\:\:{is}\:\left({Always}\:{greater}\:{than}\:\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${So} \\ $$$$\frac{\mathrm{1}}{{n}+\mathrm{1}}+\frac{\mathrm{1}}{{n}+\mathrm{2}}+\frac{\mathrm{1}}{{n}+\mathrm{3}}+….>\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Answered by 1549442205PVT last updated on 14/Oct/20
Prove that (1/(n+1))+(1/(n+2))+...+(1/(2n))>(1/2)(1) for n>2,n∈N i)For n=2 we have (1/(2+1))+(1/(2+2))=(1/3)+(1/4) =(7/(12))>(6/(12))=(1/2)⇒ The inequality is true ii)Suppose the inequality was true for n=k that means S_k =Σ_(p=1) ^(k) (1/(k+p))<(1/(2k)) iii)Need prove that S_(k+1) =Σ_(p=1) ^(k) (1/(k+1+p))<(1/(2(k+1))) Indeed,S_(k+1) =(1/(k+2))+...+(1/(2k))+(1/(2(k+1))) =(1/(k+1))+(1/(k+2))+...+(1/(2k))+((1/(2(k+1)))−(1/(k+1))) =S_k −(1/(2(k+1)))<(1/(2k))−(1/(2(k+1)))=((k+1−k)/(2k(k+1))) =(1/(2k(k+1)))<(1/(2(k+1))).This shows the inequality (1) is also true for n=k+1 By induction mathematic principle it is true ∀n∈N,n≥2 second way(don′t use induction method) Since n+k<2n ∀k=1,2,...,(n−1),so (1/(n+k))<(1/(2n ))∀k=1,2,...,(n−1).Hence (1/(n+1))+(1/(n+2))+...+(1/(2n))>(1/(2n))+(1/(2n))+...+(1/(2n)) =((1+1+...+1)/(2n))=(n/(2n))=(1/2) Since the sum consist of n terms
$$\mathrm{Prove}\:\mathrm{that}\:\:\frac{\mathrm{1}}{{n}+\mathrm{1}}+\frac{\mathrm{1}}{{n}+\mathrm{2}}+…+\frac{\mathrm{1}}{\mathrm{2}{n}}>\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}\right)\:\mathrm{for}\:\mathrm{n}>\mathrm{2},\mathrm{n}\in\mathrm{N} \\ $$$$\left.\mathrm{i}\right)\mathrm{For}\:\mathrm{n}=\mathrm{2}\:\mathrm{we}\:\mathrm{have}\:\frac{\mathrm{1}}{\mathrm{2}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}+\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$=\frac{\mathrm{7}}{\mathrm{12}}>\frac{\mathrm{6}}{\mathrm{12}}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\:\mathrm{The}\:\mathrm{inequality}\:\mathrm{is}\:\mathrm{true} \\ $$$$\left.\mathrm{ii}\right)\mathrm{Suppose}\:\mathrm{the}\:\mathrm{inequality}\:\mathrm{was}\:\mathrm{true}\: \\ $$$$\mathrm{for}\:\mathrm{n}=\mathrm{k}\:\mathrm{that}\:\mathrm{means}\:\mathrm{S}_{\mathrm{k}} \underset{\mathrm{p}=\mathrm{1}} {=\Sigma}\frac{\mathrm{1}}{\mathrm{k}+\mathrm{p}}<\frac{\mathrm{1}}{\mathrm{2k}} \\ $$$$\left.\mathrm{iii}\right)\mathrm{Need}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{S}_{\mathrm{k}+\mathrm{1}} =\underset{\mathrm{p}=\mathrm{1}} {\overset{\mathrm{k}} {\Sigma}}\:\frac{\mathrm{1}}{\mathrm{k}+\mathrm{1}+\mathrm{p}}<\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{k}+\mathrm{1}\right)} \\ $$$$\mathrm{Indeed},\mathrm{S}_{\mathrm{k}+\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{k}+\mathrm{2}}+…+\frac{\mathrm{1}}{\mathrm{2k}}+\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{k}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{k}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{k}+\mathrm{2}}+…+\frac{\mathrm{1}}{\mathrm{2k}}+\left(\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{k}+\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{k}+\mathrm{1}}\right) \\ $$$$=\mathrm{S}_{\mathrm{k}} −\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{k}+\mathrm{1}\right)}<\frac{\mathrm{1}}{\mathrm{2k}}−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{k}+\mathrm{1}\right)}=\frac{\mathrm{k}+\mathrm{1}−\mathrm{k}}{\mathrm{2k}\left(\mathrm{k}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2k}\left(\mathrm{k}+\mathrm{1}\right)}<\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{k}+\mathrm{1}\right)}.\mathrm{This}\:\mathrm{shows}\:\mathrm{the} \\ $$$$\mathrm{inequality}\:\left(\mathrm{1}\right)\:\mathrm{is}\:\mathrm{also}\:\mathrm{true}\:\mathrm{for}\:\mathrm{n}=\mathrm{k}+\mathrm{1} \\ $$$$\mathrm{By}\:\mathrm{induction}\:\mathrm{mathematic}\:\mathrm{principle} \\ $$$$\mathrm{it}\:\mathrm{is}\:\mathrm{true}\:\forall\mathrm{n}\in\mathrm{N},\mathrm{n}\geqslant\mathrm{2} \\ $$$$\boldsymbol{\mathrm{second}}\:\boldsymbol{\mathrm{way}}\left(\mathrm{don}'\mathrm{t}\:\mathrm{use}\:\mathrm{induction}\:\mathrm{method}\right) \\ $$$$\mathrm{Since}\:\mathrm{n}+\mathrm{k}<\mathrm{2n}\:\forall\mathrm{k}=\mathrm{1},\mathrm{2},…,\left(\mathrm{n}−\mathrm{1}\right),\mathrm{so} \\ $$$$\frac{\mathrm{1}}{\mathrm{n}+\mathrm{k}}<\frac{\mathrm{1}}{\mathrm{2n}\:}\forall\mathrm{k}=\mathrm{1},\mathrm{2},…,\left(\mathrm{n}−\mathrm{1}\right).\mathrm{Hence} \\ $$$$\frac{\mathrm{1}}{{n}+\mathrm{1}}+\frac{\mathrm{1}}{{n}+\mathrm{2}}+…+\frac{\mathrm{1}}{\mathrm{2}{n}}>\frac{\mathrm{1}}{\mathrm{2n}}+\frac{\mathrm{1}}{\mathrm{2n}}+…+\frac{\mathrm{1}}{\mathrm{2n}} \\ $$$$=\frac{\mathrm{1}+\mathrm{1}+…+\mathrm{1}}{\mathrm{2n}}=\frac{\mathrm{n}}{\mathrm{2n}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{Since}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{consist}\:\mathrm{of}\:\mathrm{n}\:\mathrm{terms} \\ $$

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