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Question Number 95668 by I want to learn more last updated on 26/May/20

Prove that:   2^(1/4) .4^(1/8) .8^(1/16) .16^(1/32) .  ...  ∞   =   2

$$\mathrm{Prove}\:\mathrm{that}:\:\:\:\mathrm{2}^{\mathrm{1}/\mathrm{4}} .\mathrm{4}^{\mathrm{1}/\mathrm{8}} .\mathrm{8}^{\mathrm{1}/\mathrm{16}} .\mathrm{16}^{\mathrm{1}/\mathrm{32}} .\:\:...\:\:\infty\:\:\:=\:\:\:\mathrm{2} \\ $$

Commented by Tony Lin last updated on 26/May/20

2^(1/4) ×4^(1/8) ×8^(1/(16)) ×16^(1/(32)) ×∙∙∙  =2^(1/4) ×(2^2 )^(1/8) ×(2^3 )^(1/(16)) ×(2^4 )^(1/(32)) ×∙∙∙=2^1   so what we need to prove is  (1/2^2 )+(2/2^3 )+(3/2^4 )+(4/2^5 )+∙∙∙=1  ⇒Σ_(k=1) ^∞ (k/2^(k+1) )=1  let f(x)=Σ_(k=1) ^∞ k (x^(k−1) /2^(k+1) )  let F(x)=∫_0 ^x f(t)dt=Σ_(k=1) ^∞ (x^k /2^(k+1) )+c  =(1/2)[((x/2))^1 +((x/2))^2 +((x/2))^3 +∙∙∙]+c  =(1/2)×((x/2)/(1−(x/2)))+c  =(x/(4−2x))+c  when x=0, F(x)=0  ∴ c=0  ⇒F(x)=(x/(4−2x))  f(x)=F ′(x)=(4/((4−2x)^2 ))  plug x=1  ⇒Σ_(k=1) ^∞ (k/2^(k+1) )=(4/((4−2)^2 ))=1  hence proved

$$\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} ×\mathrm{4}^{\frac{\mathrm{1}}{\mathrm{8}}} ×\mathrm{8}^{\frac{\mathrm{1}}{\mathrm{16}}} ×\mathrm{16}^{\frac{\mathrm{1}}{\mathrm{32}}} ×\centerdot\centerdot\centerdot \\ $$$$=\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} ×\left(\mathrm{2}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{8}}} ×\left(\mathrm{2}^{\mathrm{3}} \right)^{\frac{\mathrm{1}}{\mathrm{16}}} ×\left(\mathrm{2}^{\mathrm{4}} \right)^{\frac{\mathrm{1}}{\mathrm{32}}} ×\centerdot\centerdot\centerdot=\mathrm{2}^{\mathrm{1}} \\ $$$${so}\:{what}\:{we}\:{need}\:{to}\:{prove}\:{is} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{2}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{3}}{\mathrm{2}^{\mathrm{4}} }+\frac{\mathrm{4}}{\mathrm{2}^{\mathrm{5}} }+\centerdot\centerdot\centerdot=\mathrm{1} \\ $$$$\Rightarrow\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{k}}{\mathrm{2}^{{k}+\mathrm{1}} }=\mathrm{1} \\ $$$${let}\:{f}\left({x}\right)=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}{k}\:\frac{{x}^{{k}−\mathrm{1}} }{\mathrm{2}^{{k}+\mathrm{1}} } \\ $$$${let}\:{F}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} {f}\left({t}\right){dt}=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{k}} }{\mathrm{2}^{{k}+\mathrm{1}} }+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\left(\frac{{x}}{\mathrm{2}}\right)^{\mathrm{1}} +\left(\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{x}}{\mathrm{2}}\right)^{\mathrm{3}} +\centerdot\centerdot\centerdot\right]+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\frac{{x}}{\mathrm{2}}}{\mathrm{1}−\frac{{x}}{\mathrm{2}}}+{c} \\ $$$$=\frac{{x}}{\mathrm{4}−\mathrm{2}{x}}+{c} \\ $$$${when}\:{x}=\mathrm{0},\:{F}\left({x}\right)=\mathrm{0} \\ $$$$\therefore\:{c}=\mathrm{0} \\ $$$$\Rightarrow{F}\left({x}\right)=\frac{{x}}{\mathrm{4}−\mathrm{2}{x}} \\ $$$${f}\left({x}\right)={F}\:'\left({x}\right)=\frac{\mathrm{4}}{\left(\mathrm{4}−\mathrm{2}{x}\right)^{\mathrm{2}} } \\ $$$${plug}\:{x}=\mathrm{1} \\ $$$$\Rightarrow\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{k}}{\mathrm{2}^{{k}+\mathrm{1}} }=\frac{\mathrm{4}}{\left(\mathrm{4}−\mathrm{2}\right)^{\mathrm{2}} }=\mathrm{1} \\ $$$${hence}\:{proved} \\ $$

Commented by I want to learn more last updated on 06/Jun/20

Thanks sir

$$\mathrm{Thanks}\:\mathrm{sir} \\ $$

Answered by Rio Michael last updated on 26/May/20

my try please check.   2^(1/4) .4^(1/8) .8^(1/16) .16^(1/32) ...∞ = Π_(n=1) ^∞ 2^(n/(2^(n+1)  ))    now let x = Π_(n=1) ^∞ 2^(n/2^(n+1) )    ⇔ ln x = Σ_(n=1) ^∞ (n/2^(n+1) ) ln 2  ..........(i)     ln x = ln 2 Σ_(n=1) ^∞ (n/2^(n+1) ) = (1/2^2 ) + (2/2^3 ) + (3/2^4 ) + ...     = (1/2^2 ) +(1/2^3 ) + (1/2^4 ) + .... is a GP with r = (1/2) ⇒ S_∞  = (1/(16))  ⇒ Σ_(n=1) ^∞ 2^(n/(n+1))  = (1/2) + (1/4) + (1/8) + ...                          = (1/2)(1 + (1/2) + (1/4) + ...) = (1/2)(1/(1−(1/2))) = 1 .........(ii)   eqn (i) + eqn(ii)  ⇒ ln x = ln 2 × 1 = ln 2 ⇔ x = 2   ∴ Π_(n=1) ^∞ 2^(n/2^(n+1) )  = 2^(1/4) .4^(1/8) .8^(1/16) .16^(1/32) ....∞ = 2.

$$\mathrm{my}\:\mathrm{try}\:\mathrm{please}\:\mathrm{check}. \\ $$$$\:\mathrm{2}^{\mathrm{1}/\mathrm{4}} .\mathrm{4}^{\mathrm{1}/\mathrm{8}} .\mathrm{8}^{\mathrm{1}/\mathrm{16}} .\mathrm{16}^{\mathrm{1}/\mathrm{32}} ...\infty\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\mathrm{2}^{\frac{{n}}{\mathrm{2}^{{n}+\mathrm{1}} \:}} \: \\ $$$$\mathrm{now}\:\mathrm{let}\:{x}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\mathrm{2}^{\frac{{n}}{\mathrm{2}^{{n}+\mathrm{1}} }} \: \\ $$$$\Leftrightarrow\:\mathrm{ln}\:{x}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}}{\mathrm{2}^{{n}+\mathrm{1}} }\:\mathrm{ln}\:\mathrm{2}\:\:..........\left({i}\right) \\ $$$$\:\:\:\mathrm{ln}\:{x}\:=\:\mathrm{ln}\:\mathrm{2}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}}{\mathrm{2}^{{n}+\mathrm{1}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\:+\:\frac{\mathrm{2}}{\mathrm{2}^{\mathrm{3}} }\:+\:\frac{\mathrm{3}}{\mathrm{2}^{\mathrm{4}} }\:+\:... \\ $$$$\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }\:+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{4}} }\:+\:....\:\mathrm{is}\:\mathrm{a}\:\mathrm{GP}\:\mathrm{with}\:{r}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:{S}_{\infty} \:=\:\frac{\mathrm{1}}{\mathrm{16}} \\ $$$$\Rightarrow\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{2}^{\frac{{n}}{{n}+\mathrm{1}}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{4}}\:+\:\frac{\mathrm{1}}{\mathrm{8}}\:+\:...\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{4}}\:+\:...\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}\:=\:\mathrm{1}\:.........\left({ii}\right) \\ $$$$\:\mathrm{eqn}\:\left({i}\right)\:+\:\mathrm{eqn}\left({ii}\right)\:\:\Rightarrow\:\mathrm{ln}\:{x}\:=\:\mathrm{ln}\:\mathrm{2}\:×\:\mathrm{1}\:=\:\mathrm{ln}\:\mathrm{2}\:\Leftrightarrow\:{x}\:=\:\mathrm{2} \\ $$$$\:\therefore\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\mathrm{2}^{\frac{{n}}{\mathrm{2}^{{n}+\mathrm{1}} }} \:=\:\mathrm{2}^{\mathrm{1}/\mathrm{4}} .\mathrm{4}^{\mathrm{1}/\mathrm{8}} .\mathrm{8}^{\mathrm{1}/\mathrm{16}} .\mathrm{16}^{\mathrm{1}/\mathrm{32}} ....\infty\:=\:\mathrm{2}. \\ $$

Commented by I want to learn more last updated on 06/Jun/20

Thanks sir

$$\mathrm{Thanks}\:\mathrm{sir} \\ $$

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