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Question Number 91640 by mathmax by abdo last updated on 02/May/20

find nature of the serie Σ_n  cos(πn^2 ln(1+(1/n)))

$${find}\:{nature}\:{of}\:{the}\:{serie}\:\sum_{{n}} \:{cos}\left(\pi{n}^{\mathrm{2}} {ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)\right) \\ $$

Commented by mathmax by abdo last updated on 04/May/20

we have ln^′ (1+u) =(1/(1+u)) =1−u +u^2 +o(u^2 )   (∣u∣<1) ⇒  ln(1+u) =u−(u^2 /2)+(u^3 /3)o(u^2 ) ⇒ln(1+(1/n))=(1/n)−(1/(2n^2 )) +(1/(3n^3 ))+o((1/n^3 )) ⇒  πn^2 ln(1+(1/n)) =πn−(π/2) +(π/(3n)) +o((1/n)) ⇒  cos(πn^2 ln(1+(1/n))) ∼cos(πn−(π/2)+(π/(3n)))  =cos((π/2)−(πn +(π/(3n)))) =sin(nπ +(π/(3n))) =(−1)^n  sin((π/(3n)))∼(−1)^n  (π/(3n))  let v_n =(π/(3n))  (v_n )decrease to 0 ⇒Σ(−1)^n  v_n  is convergent   (alternate serie) ⇒Σ_n cos(πn^2 ln(1+(1/n))) converges ...!

$${we}\:{have}\:{ln}^{'} \left(\mathrm{1}+{u}\right)\:=\frac{\mathrm{1}}{\mathrm{1}+{u}}\:=\mathrm{1}−{u}\:+{u}^{\mathrm{2}} +{o}\left({u}^{\mathrm{2}} \right)\:\:\:\left(\mid{u}\mid<\mathrm{1}\right)\:\Rightarrow \\ $$$${ln}\left(\mathrm{1}+{u}\right)\:={u}−\frac{{u}^{\mathrm{2}} }{\mathrm{2}}+\frac{{u}^{\mathrm{3}} }{\mathrm{3}}{o}\left({u}^{\mathrm{2}} \right)\:\Rightarrow{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)=\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{3}{n}^{\mathrm{3}} }+{o}\left(\frac{\mathrm{1}}{{n}^{\mathrm{3}} }\right)\:\Rightarrow \\ $$$$\pi{n}^{\mathrm{2}} {ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)\:=\pi{n}−\frac{\pi}{\mathrm{2}}\:+\frac{\pi}{\mathrm{3}{n}}\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\:\Rightarrow \\ $$$${cos}\left(\pi{n}^{\mathrm{2}} {ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)\right)\:\sim{cos}\left(\pi{n}−\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{3}{n}}\right) \\ $$$$={cos}\left(\frac{\pi}{\mathrm{2}}−\left(\pi{n}\:+\frac{\pi}{\mathrm{3}{n}}\right)\right)\:={sin}\left({n}\pi\:+\frac{\pi}{\mathrm{3}{n}}\right)\:=\left(−\mathrm{1}\right)^{{n}} \:{sin}\left(\frac{\pi}{\mathrm{3}{n}}\right)\sim\left(−\mathrm{1}\right)^{{n}} \:\frac{\pi}{\mathrm{3}{n}} \\ $$$${let}\:{v}_{{n}} =\frac{\pi}{\mathrm{3}{n}}\:\:\left({v}_{{n}} \right){decrease}\:{to}\:\mathrm{0}\:\Rightarrow\Sigma\left(−\mathrm{1}\right)^{{n}} \:{v}_{{n}} \:{is}\:{convergent}\: \\ $$$$\left({alternate}\:{serie}\right)\:\Rightarrow\sum_{{n}} {cos}\left(\pi{n}^{\mathrm{2}} {ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)\right)\:{converges}\:...! \\ $$

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