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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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