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Question Number 82452 by mathmax by abdo last updated on 21/Feb/20

nature ofthe serie Σ_(n=0) ^∞ ln(cos((1/2^n )))

$${nature}\:{ofthe}\:{serie}\:\sum_{{n}=\mathrm{0}} ^{\infty} {ln}\left({cos}\left(\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\right)\right) \\ $$

Commented by mathmax by abdo last updated on 23/Feb/20

cos((1/2^n ))∼1−((((1/2^n ))^2 )/2) =1−(1/2^(2n+1) ) and ln(cos((1/2^n )))∼ln(1−(1/2^(2n+1) )) ∼−(1/2^(2n+1) ) and Σ_(n=0) ^∞ (1/2^(2n+1) ) converges ⇒Σ_(n=0) ^∞ ln(cos((1/2^n ))) is convergente.

$${cos}\left(\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\right)\sim\mathrm{1}−\frac{\left(\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\right)^{\mathrm{2}} }{\mathrm{2}}\:=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} }\:{and}\:{ln}\left({cos}\left(\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\right)\right)\sim{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} }\right) \\ $$$$\sim−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} }\:\:{and}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} }\:{converges}\:\Rightarrow\sum_{{n}=\mathrm{0}} ^{\infty} {ln}\left({cos}\left(\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\right)\right) \\ $$$${is}\:{convergente}. \\ $$

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