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Question Number 64561 by aliesam last updated on 19/Jul/19

Commented by mathmax by abdo last updated on 20/Jul/19

let P_n =Π_(k=0) ^n (1+σ^2^k ) ⇒P_n =(1+σ)(1+σ^2 )(1+σ^4 )...(1+σ^2^n ) we can prove by recurence that (1+σ^2 )(1+σ^4 )...(1+σ^2^n )=((1−σ^2^(n+1) )/(1−σ^2 )) if σ^2 ≠1( here 0<σ<1) ⇒P_n =(1+σ)((1−σ^2^(n+1) )/(1−σ^2 )) but lim_(n→+∞) σ^2^(n+1) =0 ⇒lim_(n→+∞) P_n =((1+σ)/(1−σ^2 )) =(1/(1−σ)) finally lim_(n→+∞) Π_(k=0) ^n (1+σ^2^k ) =(1/(1−σ)) .

$${let}\:{P}_{{n}} =\prod_{{k}=\mathrm{0}} ^{{n}} \left(\mathrm{1}+\sigma^{\mathrm{2}^{{k}} } \right)\:\Rightarrow{P}_{{n}} =\left(\mathrm{1}+\sigma\right)\left(\mathrm{1}+\sigma^{\mathrm{2}} \right)\left(\mathrm{1}+\sigma^{\mathrm{4}} \right)...\left(\mathrm{1}+\sigma^{\mathrm{2}^{{n}} } \right) \\ $$$${we}\:{can}\:{prove}\:{by}\:{recurence}\:{that}\:\left(\mathrm{1}+\sigma^{\mathrm{2}} \right)\left(\mathrm{1}+\sigma^{\mathrm{4}} \right)...\left(\mathrm{1}+\sigma^{\mathrm{2}^{{n}} } \right)=\frac{\mathrm{1}−\sigma^{\mathrm{2}^{{n}+\mathrm{1}} } }{\mathrm{1}−\sigma^{\mathrm{2}} } \\ $$$${if}\:\sigma^{\mathrm{2}} \neq\mathrm{1}\left(\:{here}\:\:\mathrm{0}<\sigma<\mathrm{1}\right)\:\Rightarrow{P}_{{n}} =\left(\mathrm{1}+\sigma\right)\frac{\mathrm{1}−\sigma^{\mathrm{2}^{{n}+\mathrm{1}} } }{\mathrm{1}−\sigma^{\mathrm{2}} } \\ $$$${but}\:{lim}_{{n}\rightarrow+\infty} \:\sigma^{\mathrm{2}^{{n}+\mathrm{1}} } \:=\mathrm{0}\:\Rightarrow{lim}_{{n}\rightarrow+\infty} \:{P}_{{n}} =\frac{\mathrm{1}+\sigma}{\mathrm{1}−\sigma^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{1}−\sigma} \\ $$$${finally}\:\:{lim}_{{n}\rightarrow+\infty} \:\prod_{{k}=\mathrm{0}} ^{{n}} \left(\mathrm{1}+\sigma^{\mathrm{2}^{{k}} } \right)\:=\frac{\mathrm{1}}{\mathrm{1}−\sigma}\:. \\ $$

Answered by JDamian last updated on 20/Jul/19

The answer is (1/(1−δ)) let P_n =Π_(k=0) ^(n) (1+δ^2^k ). It is easy to show that P_n is actually the sum of the following geometric progession, which: − first value is 1. − common ratio is δ. − length is 2^n . P_n =Σ_(k=0) ^(2^n −1) δ^k ∀δ 0<δ<1 lim_(n→∞) P_n =(1/(1−δ))

$${The}\:{answer}\:{is}\:\frac{\mathrm{1}}{\mathrm{1}−\delta} \\ $$$$ \\ $$$${let}\:{P}_{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\Pi}}\left(\mathrm{1}+\delta^{\mathrm{2}^{{k}} } \right).\:{It}\:{is}\:{easy}\:{to}\:{show}\:{that} \\ $$$${P}_{{n}} \:{is}\:{actually}\:{the}\:\boldsymbol{{sum}}\:\boldsymbol{{of}}\:\boldsymbol{{the}}\:\boldsymbol{{following}} \\ $$$$\boldsymbol{{geometric}}\:\boldsymbol{{progession}},\:{which}: \\ $$$$−\:{first}\:{value}\:{is}\:\mathrm{1}. \\ $$$$−\:{common}\:{ratio}\:{is}\:\delta. \\ $$$$−\:{length}\:{is}\:\mathrm{2}^{{n}} . \\ $$$$ \\ $$$${P}_{{n}} =\underset{{k}=\mathrm{0}} {\overset{\mathrm{2}^{{n}} −\mathrm{1}} {\sum}}\delta^{{k}} \:\:\forall\delta\:\:\:\mathrm{0}<\delta<\mathrm{1} \\ $$$$\underset{{n}\rightarrow\infty} {{lim}}\:{P}_{{n}} =\frac{\mathrm{1}}{\mathrm{1}−\delta} \\ $$

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