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Question Number 108371 by Skabetix last updated on 16/Aug/20
prove by reccurence (1/2)+(1/4)+(1/6)+...+(1/(2n))<or =(n/2) thanks
$${prove}\:{by}\:{reccurence}\: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{6}}+…+\frac{\mathrm{1}}{\mathrm{2}{n}}<{or}\:=\frac{{n}}{\mathrm{2}} \\ $$$${thanks} \\ $$
Answered by Aziztisffola last updated on 16/Aug/20
for n=1 ⇒(1/2)≤(1/2) true suppose (1/2)+(1/4)+(1/6)+...+(1/(2n))≤(n/2) and prove that (1/2)+(1/4)+(1/6)+...+(1/(2n))+(1/(2n+2))≤((n+1)/2) (1/2)+(1/4)+(1/6)+...+(1/(2n))+(1/(2n+2))≤(n/2)+(1/(2n+2)) (n/2)+(1/(2n+2))=(n/2)+(1/(2(n+1)))≤(n/2)+(1/2) =((n+1)/2) (1/2)+(1/4)+(1/6)+...+(1/(2n))+(1/(2n+2))≤((n+1)/2) hence ∀n∈N^∗ Σ_(k=1) ^n (1/(2k))≤(n/2)
$$\:\mathrm{for}\:\mathrm{n}=\mathrm{1}\:\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\leqslant\frac{\mathrm{1}}{\mathrm{2}}\:\:\mathrm{true} \\ $$$$\:\mathrm{suppose}\:\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{6}}+…+\frac{\mathrm{1}}{\mathrm{2}{n}}\leqslant\frac{\mathrm{n}}{\mathrm{2}} \\ $$$$\:\mathrm{and}\:\mathrm{prove}\:\mathrm{that}\:\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{6}}+…+\frac{\mathrm{1}}{\mathrm{2}{n}}+\frac{\mathrm{1}}{\mathrm{2n}+\mathrm{2}}\leqslant\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{6}}+…+\frac{\mathrm{1}}{\mathrm{2}{n}}+\frac{\mathrm{1}}{\mathrm{2n}+\mathrm{2}}\leqslant\frac{\mathrm{n}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2n}+\mathrm{2}} \\ $$$$\frac{\mathrm{n}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2n}+\mathrm{2}}=\frac{\mathrm{n}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{n}+\mathrm{1}\right)}\leqslant\frac{\mathrm{n}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\:\:=\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{6}}+…+\frac{\mathrm{1}}{\mathrm{2}{n}}+\frac{\mathrm{1}}{\mathrm{2n}+\mathrm{2}}\leqslant\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{hence}\:\forall\mathrm{n}\in\mathbb{N}^{\ast} \:\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{2k}}\leqslant\frac{\mathrm{n}}{\mathrm{2}} \\ $$
Answered by 1549442205PVT last updated on 16/Aug/20
i)For n=1 we have (1/2)=(1/2)⇒true ii)Consider n≥5 we have (1/(10))<(1/(4.5))=(1/4)−(1/5),(1/(12))<(1/(5.6))=(1/5)−(1/6)... (1/(2n))<(1/((n−1)n))=(1/(n−1))−(1/n) ⇒(1/(10))+(1/(12))+...+(1/(2n))<(1/4)−(1/5)+(1/5)−(1/6)+... ...+(1/(n−1))−(1/n)=(1/4)−(1/n).Hence, Adding up we get LHS<((1/2)+(1/4)+(1/6)+(1/8))+(1/4)−(1/n)<(n/2) ⇔((12+6+4+3)/(24))−(1/n)<(n/2)⇔((25)/(24))−(1/n)<(n/2) ⇔((25n−24)/(24n))<(n/2)⇔12n^2 −25n+24>0 ⇔12(n−((25)/(24)))^2 +24−((625)/(48))>0 is true ∀n≥2 Consequently, (1/2)+(1/4)+(1/6)+...+(1/(2n))≤(n/2)(q.e.d)
$$\left.\mathrm{i}\right)\mathrm{For}\:\mathrm{n}=\mathrm{1}\:\mathrm{we}\:\mathrm{have}\:\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{true} \\ $$$$\left.\mathrm{ii}\right)\mathrm{Consider}\:\mathrm{n}\geqslant\mathrm{5}\:\mathrm{we}\:\mathrm{have} \\ $$$$\frac{\mathrm{1}}{\mathrm{10}}<\frac{\mathrm{1}}{\mathrm{4}.\mathrm{5}}=\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{5}},\frac{\mathrm{1}}{\mathrm{12}}<\frac{\mathrm{1}}{\mathrm{5}.\mathrm{6}}=\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{6}}… \\ $$$$\frac{\mathrm{1}}{\mathrm{2n}}<\frac{\mathrm{1}}{\left(\mathrm{n}−\mathrm{1}\right)\mathrm{n}}=\frac{\mathrm{1}}{\mathrm{n}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{n}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{10}}+\frac{\mathrm{1}}{\mathrm{12}}+…+\frac{\mathrm{1}}{\mathrm{2n}}<\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{6}}+… \\ $$$$…+\frac{\mathrm{1}}{\mathrm{n}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{n}}=\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{n}}.\mathrm{Hence}, \\ $$$$\mathrm{Adding}\:\mathrm{up}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{LHS}<\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{8}}\right)+\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{n}}<\frac{\mathrm{n}}{\mathrm{2}} \\ $$$$\Leftrightarrow\frac{\mathrm{12}+\mathrm{6}+\mathrm{4}+\mathrm{3}}{\mathrm{24}}−\frac{\mathrm{1}}{\mathrm{n}}<\frac{\mathrm{n}}{\mathrm{2}}\Leftrightarrow\frac{\mathrm{25}}{\mathrm{24}}−\frac{\mathrm{1}}{\mathrm{n}}<\frac{\mathrm{n}}{\mathrm{2}} \\ $$$$\Leftrightarrow\frac{\mathrm{25n}−\mathrm{24}}{\mathrm{24n}}<\frac{\mathrm{n}}{\mathrm{2}}\Leftrightarrow\mathrm{12n}^{\mathrm{2}} −\mathrm{25n}+\mathrm{24}>\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{12}\left(\mathrm{n}−\frac{\mathrm{25}}{\mathrm{24}}\right)^{\mathrm{2}} +\mathrm{24}−\frac{\mathrm{625}}{\mathrm{48}}>\mathrm{0} \\ $$$$\mathrm{is}\:\mathrm{true}\:\forall\mathrm{n}\geqslant\mathrm{2} \\ $$$$\mathrm{Consequently}, \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{6}}+…+\frac{\mathrm{1}}{\mathrm{2}\boldsymbol{{n}}}\leqslant\frac{\boldsymbol{\mathrm{n}}}{\mathrm{2}}\left(\boldsymbol{\mathrm{q}}.\boldsymbol{\mathrm{e}}.\boldsymbol{\mathrm{d}}\right) \\ $$$$ \\ $$

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