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Question Number 141132 by mnjuly1970 last updated on 16/May/21
      ...mathematical ...analysis...      prove  that:           Π_(n=0) ^∞ (1+(1/2^2^n  ) ) =^?  2           ......
$$\:\:\: \\ $$$$\:…{mathematical}\:…{analysis}… \\ $$$$\:\:\:\:{prove}\:\:{that}: \\ $$$$\:\:\:\:\:\:\:\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\prod}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}^{{n}} } }\:\right)\:\overset{?} {=}\:\mathrm{2}\:\:\: \\ $$$$\:\:\:\:\:\:…… \\ $$
Answered by Dwaipayan Shikari last updated on 16/May/21
Π_(n=0) ^∞ (1+(1/x^2^n  ))=Π_(n=0) ^∞ ((1−(1/x^2^(n+1)  ))/(1−(1/x^2^n  )))=((1−(1/x^2 ))/(1−(1/x))).((1−(1/x^4 ))/(1−(1/x^2 ))).((1−(1/x^8 ))/(1−(1/x^4 ))).((1−(1/x^(16) ))/(1−(1/x^8 )))...((1−(1/x^2^(N+1)  ))/(1−(1/x^2^N  )))  =lim_(N→∞) (x/(x−1)).(1−(1/x^2^(N+1)  ))=(x/(x−1))  x=2
$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\prod}}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}^{{n}} } }\right)=\underset{{n}=\mathrm{0}} {\overset{\infty} {\prod}}\frac{\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}^{{n}+\mathrm{1}} } }}{\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}^{{n}} } }}=\frac{\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{\mathrm{1}−\frac{\mathrm{1}}{{x}}}.\frac{\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{4}} }}{\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}.\frac{\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{8}} }}{\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{4}} }}.\frac{\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{16}} }}{\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{8}} }}…\frac{\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}^{{N}+\mathrm{1}} } }}{\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}^{{N}} } }} \\ $$$$=\underset{{N}\rightarrow\infty} {\mathrm{lim}}\frac{{x}}{{x}−\mathrm{1}}.\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}^{{N}+\mathrm{1}} } }\right)=\frac{{x}}{{x}−\mathrm{1}} \\ $$$${x}=\mathrm{2} \\ $$
Commented by mnjuly1970 last updated on 16/May/21
very nice..
$${very}\:{nice}.. \\ $$

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