Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

Question Number 118488 by Lordose last updated on 18/Oct/20

Conjecture a formula for the infinite sum of the series. (1/3)+(1/(15))+(1/(35))+ ∙ ∙ ∙ (1/((2n−1)(2n+1))) And prove the formula by Induction.

$$\boldsymbol{\mathrm{Conjecture}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{formula}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{infinite}} \\ $$$$\boldsymbol{\mathrm{sum}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{series}}. \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{15}}+\frac{\mathrm{1}}{\mathrm{35}}+\:\centerdot\:\centerdot\:\centerdot\:\frac{\mathrm{1}}{\left(\mathrm{2}\boldsymbol{\mathrm{n}}−\mathrm{1}\right)\left(\mathrm{2}\boldsymbol{\mathrm{n}}+\mathrm{1}\right)} \\ $$$$\boldsymbol{\mathrm{And}}\:\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{formula}}\:\boldsymbol{\mathrm{by}}\:\boldsymbol{\mathrm{Induction}}. \\ $$

Answered by Olaf last updated on 18/Oct/20

(1/((2k−1)(2k+1))) = (1/2)((1/(2k−1))−(1/(2k+1))) S_n = Σ_(k=1) ^n (1/((2k−1)(2k+1))) S_n = (1/2)Σ_(k=1) ^n (1/(2k−1))−(1/2)Σ_(k=1) ^n (1/(2k+1)) S_n = (1/2)Σ_(k=0) ^(n−1) (1/(2k+1))−(1/2)Σ_(k=1) ^n (1/(2k+1)) S_n = (1/2)(1+Σ_(k=1) ^n (1/(2k+1))−(1/(2n+1)))−(1/2)Σ_(k=1) ^n (1/(2k+1)) S_n = (1/2)(1−(1/(2n+1))) = (n/(2n+1)) Induction : for n = 1, S_1 = (1/(1.3)) = (1/3) = (1/(2(1)+1)) ⇒ the formula is true for n = 1 Now we suppose the formula is true for n. S_(n+1) = S_n +(1/((2n+1)(2n+3))) S_(n+1) = (n/(2n+1))+(1/((2n+1)(2n+3))) S_(n+1) = ((n(2n+3)+1)/((2n+1)(2n+3))) S_(n+1) = ((2n^2 +3n+1)/((2n+1)(2n+3))) S_(n+1) = ((2(n+1)(n+(1/2)))/((2n+1)(2n+3))) S_(n+1) = ((n+1)/(2n+3)) S_n true ⇒ S_(n+1) true Finally, S_n = (n/(2n+1)), n ≥ 1

$$\frac{\mathrm{1}}{\left(\mathrm{2}{k}−\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{1}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}\right) \\ $$$$\mathrm{S}_{{n}} \:=\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{k}−\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{1}\right)} \\ $$$$\mathrm{S}_{{n}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}} \\ $$$$\mathrm{S}_{{n}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}} \\ $$$$\mathrm{S}_{{n}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}} \\ $$$$\mathrm{S}_{{n}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\right)\:=\:\frac{{n}}{\mathrm{2}{n}+\mathrm{1}} \\ $$$$ \\ $$$$\mathrm{Induction}\:: \\ $$$$\mathrm{for}\:{n}\:=\:\mathrm{1},\:\mathrm{S}_{\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{1}.\mathrm{3}}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\:=\:\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}\right)+\mathrm{1}} \\ $$$$\Rightarrow\:\mathrm{the}\:\mathrm{formula}\:\mathrm{is}\:\mathrm{true}\:\mathrm{for}\:{n}\:=\:\mathrm{1} \\ $$$$ \\ $$$$\mathrm{Now}\:\mathrm{we}\:\mathrm{suppose}\:\mathrm{the}\:\mathrm{formula}\:\mathrm{is}\:\mathrm{true}\:\mathrm{for}\:{n}. \\ $$$$\mathrm{S}_{{n}+\mathrm{1}} \:=\:\mathrm{S}_{{n}} +\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)} \\ $$$$\mathrm{S}_{{n}+\mathrm{1}} \:=\:\frac{{n}}{\mathrm{2}{n}+\mathrm{1}}+\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)} \\ $$$$\mathrm{S}_{{n}+\mathrm{1}} \:=\:\frac{{n}\left(\mathrm{2}{n}+\mathrm{3}\right)+\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)} \\ $$$$\mathrm{S}_{{n}+\mathrm{1}} \:=\:\frac{\mathrm{2}{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)} \\ $$$$\mathrm{S}_{{n}+\mathrm{1}} \:=\:\frac{\mathrm{2}\left({n}+\mathrm{1}\right)\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)} \\ $$$$\mathrm{S}_{{n}+\mathrm{1}} \:=\:\frac{{n}+\mathrm{1}}{\mathrm{2}{n}+\mathrm{3}} \\ $$$$\mathrm{S}_{{n}} \:\mathrm{true}\:\Rightarrow\:\mathrm{S}_{{n}+\mathrm{1}} \:\mathrm{true} \\ $$$$ \\ $$$$\mathrm{Finally},\:\mathrm{S}_{{n}} \:=\:\frac{{n}}{\mathrm{2}{n}+\mathrm{1}},\:{n}\:\geqslant\:\mathrm{1} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com