Question Number 91640 by mathmax by abdo last updated on 02/May/20
$${find}\:{nature}\:{of}\:{the}\:{serie}\:\sum_{{n}} \:{cos}\left(\pi{n}^{\mathrm{2}} {ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)\right) \\ $$
Commented by mathmax by abdo last updated on 04/May/20
$${we}\:{have}\:{ln}^{'} \left(\mathrm{1}+{u}\right)\:=\frac{\mathrm{1}}{\mathrm{1}+{u}}\:=\mathrm{1}−{u}\:+{u}^{\mathrm{2}} +{o}\left({u}^{\mathrm{2}} \right)\:\:\:\left(\mid{u}\mid<\mathrm{1}\right)\:\Rightarrow \\ $$$${ln}\left(\mathrm{1}+{u}\right)\:={u}−\frac{{u}^{\mathrm{2}} }{\mathrm{2}}+\frac{{u}^{\mathrm{3}} }{\mathrm{3}}{o}\left({u}^{\mathrm{2}} \right)\:\Rightarrow{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)=\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{3}{n}^{\mathrm{3}} }+{o}\left(\frac{\mathrm{1}}{{n}^{\mathrm{3}} }\right)\:\Rightarrow \\ $$$$\pi{n}^{\mathrm{2}} {ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)\:=\pi{n}−\frac{\pi}{\mathrm{2}}\:+\frac{\pi}{\mathrm{3}{n}}\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\:\Rightarrow \\ $$$${cos}\left(\pi{n}^{\mathrm{2}} {ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)\right)\:\sim{cos}\left(\pi{n}−\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{3}{n}}\right) \\ $$$$={cos}\left(\frac{\pi}{\mathrm{2}}−\left(\pi{n}\:+\frac{\pi}{\mathrm{3}{n}}\right)\right)\:={sin}\left({n}\pi\:+\frac{\pi}{\mathrm{3}{n}}\right)\:=\left(−\mathrm{1}\right)^{{n}} \:{sin}\left(\frac{\pi}{\mathrm{3}{n}}\right)\sim\left(−\mathrm{1}\right)^{{n}} \:\frac{\pi}{\mathrm{3}{n}} \\ $$$${let}\:{v}_{{n}} =\frac{\pi}{\mathrm{3}{n}}\:\:\left({v}_{{n}} \right){decrease}\:{to}\:\mathrm{0}\:\Rightarrow\Sigma\left(−\mathrm{1}\right)^{{n}} \:{v}_{{n}} \:{is}\:{convergent}\: \\ $$$$\left({alternate}\:{serie}\right)\:\Rightarrow\sum_{{n}} {cos}\left(\pi{n}^{\mathrm{2}} {ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)\right)\:{converges}\:…! \\ $$