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Question Number 90637 by Tony Lin last updated on 25/Apr/20

Σ_(n=1) ^∞ (1/2^n )tan((1/2^n ))

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{{n}} }{tan}\left(\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\right) \\ $$

Answered by TANMAY PANACEA. last updated on 25/Apr/20

tanθ−cotθ =((tan^2 θ−1)/(tanθ)) =−2(((1−tan^2 θ)/(2tanθ)))=((−2)/(tan2θ)) tanθ=cotθ−2cot2θ S=(1/2)tan((1/2))+(1/2^2 )tan((1/2^2 ))+(1/2^3 )tan((1/2^3 ))+...∞ T_1 =(1/2)tan((1/2))=(1/2)cot((1/2))−(1/2)×2cot(2×(1/2)) T_2 =(1/2^2 )tan((1/2^2 ))=(1/2^2 )cot((1/2^2 ))−(1/2^2 )×2cot(2×(1/2^2 )) T_3 =(1/2^3 )tan((1/2^3 ))=(1/2^3 )cot((1/2^3 ))−(1/2^3 )×2cot(2×(1/2^3 )) .... .... T_n =(1/2^n )tan((1/2^n ))=(1/2^n )cot((1/2^n ))−(1/2^n )×2cot(2×(1/2^n )) now add them S_n =(1/2^n )cot((1/2^n ))−(1/2)×2cot(2×(1/2))

$${tan}\theta−{cot}\theta \\ $$$$=\frac{{tan}^{\mathrm{2}} \theta−\mathrm{1}}{{tan}\theta} \\ $$$$=−\mathrm{2}\left(\frac{\mathrm{1}−{tan}^{\mathrm{2}} \theta}{\mathrm{2}{tan}\theta}\right)=\frac{−\mathrm{2}}{{tan}\mathrm{2}\theta} \\ $$$${tan}\theta={cot}\theta−\mathrm{2}{cot}\mathrm{2}\theta \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{2}}{tan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }{tan}\left(\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\right)+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }{tan}\left(\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }\right)+...\infty \\ $$$${T}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}{tan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}}{cot}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}{cot}\left(\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${T}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }{tan}\left(\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\right)=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }{cot}\left(\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\right)−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }×\mathrm{2}{cot}\left(\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\right) \\ $$$${T}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }{tan}\left(\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }\right)=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }{cot}\left(\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }\right)−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }×\mathrm{2}{cot}\left(\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }\right) \\ $$$$.... \\ $$$$.... \\ $$$${T}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}^{{n}} }{tan}\left(\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\right)=\frac{\mathrm{1}}{\mathrm{2}^{{n}} }{cot}\left(\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\right)−\frac{\mathrm{1}}{\mathrm{2}^{{n}} }×\mathrm{2}{cot}\left(\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\right) \\ $$$${now}\:{add}\:{them} \\ $$$${S}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}^{{n}} }{cot}\left(\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\right)−\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}{cot}\left(\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$ \\ $$

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

thanks sir

$${thanks}\:{sir} \\ $$

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

then lim_(n→∞) ((cot((1/2^n )))/2^n ) =lim_(n→∞) ((cos((1/2^n )))/(2^n sin((1/2^n )))) =lim_(n→∞) ((1/2^n )/(sin((1/2^n )))) let (1/2^n )=t,n→∞ ,t→0 ⇒lim_(t→0) (t/(sint)) =1 therefore, Σ_(n=1) ^∞ (1/2^n )tan((1/2^n )) =1−cot1 ≈0.3579

$${then} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{cot}\left(\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\right)}{\mathrm{2}^{{n}} } \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{cos}\left(\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\right)}{\mathrm{2}^{{n}} {sin}\left(\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\right)} \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\frac{\mathrm{1}}{\mathrm{2}^{{n}} }}{{sin}\left(\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\right)} \\ $$$${let}\:\frac{\mathrm{1}}{\mathrm{2}^{{n}} }={t},{n}\rightarrow\infty\:,{t}\rightarrow\mathrm{0} \\ $$$$\Rightarrow\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{t}}{{sint}} \\ $$$$=\mathrm{1} \\ $$$${therefore}, \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{{n}} }{tan}\left(\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\right) \\ $$$$=\mathrm{1}−{cot}\mathrm{1} \\ $$$$\approx\mathrm{0}.\mathrm{3579} \\ $$

Commented by TANMAY PANACEA. last updated on 26/Apr/20

thank you sir ...where i stopped...you helped to reach the goal...

$${thank}\:{you}\:{sir}\:...{where}\:{i}\:{stopped}...{you}\:{helped}\: \\ $$$${to}\:{reach}\:{the}\:{goal}... \\ $$

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