Question Number 159670 by Ar Brandon last updated on 19/Nov/21
Answered by puissant last updated on 20/Nov/21
$$\left.\mathrm{1}\right) \\ $$$${U}_{{n}} =\frac{{n}}{{n}^{\mathrm{3}} +\mathrm{1}}\:\underset{+\infty} {\sim}\frac{{n}}{{n}^{\mathrm{3}} }\:=\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:\:{ainsi},\:{si}\:{on}\:{compare} \\ $$$${la}\:{serie}\:{de}\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:{a}\:{une}\:{serie}\:{de}\:{riemann}\:{qui}\:{est}\: \\ $$$${convergente},\:{on}\:{remarque}\:{que}\:{cette}\:{serie}\: \\ $$$${converge}\:{egalement}.. \\ $$$$\left.\mathrm{2}\right) \\ $$$${U}_{{n}} =\:\frac{\sqrt{{n}}}{{n}^{\mathrm{2}} +\sqrt{{n}}}\:\underset{+\infty} {\sim}\frac{\sqrt{{n}}}{{n}^{\mathrm{2}} }\:\sim\:\frac{\mathrm{1}}{{n}^{\frac{\mathrm{3}}{\mathrm{2}}} }\:\:{ainsi}\:{cette}\: \\ $$$${serie}\:{est}\:{convergente}\:{car}\:{elle}\:{est}\: \\ $$$${comparable}\:{a}\:{une}\:{serie}\:{de}\:{Riemann}\: \\ $$$${qui}\:{converge}… \\ $$$$\left.\mathrm{3}\right) \\ $$$${U}_{{n}} ={nsin}\left(\frac{\mathrm{1}}{{n}}\right)\:{en}\:{calculant}\:{la}\:{limite}, \\ $$$${on}\:{a}: \\ $$$$\underset{{n}\rightarrow+\infty} {\mathrm{lim}}{nsin}\left(\frac{\mathrm{1}}{{n}}\right)\:=\:\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\frac{{sin}\left(\frac{\mathrm{1}}{{n}}\right)}{\frac{\mathrm{1}}{{n}}}\:=\:\mathrm{1} \\ $$$${et}\:{la}\:{serie}\:{la}\:{serie}\:{diverge}\:{grossi}\grave {{e}rement}.. \\ $$$$\left.\mathrm{4}\right) \\ $$$${U}_{{n}} =\frac{\mathrm{1}}{\:\sqrt{{n}}}{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{{n}}}\right) \\ $$$${En}\:{effet}\:,\:{on}\:{a}:\:{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{{n}}}\right)\:\underset{+\infty} {\sim}\frac{\mathrm{1}}{\:\sqrt{{n}}} \\ $$$${Donc}\:{U}_{{n}} \:\underset{+\infty} {\sim}\frac{\mathrm{1}}{{n}}\:{qui}\:{est}\:{donc}\:{divergente} \\ $$$${par}\:{comparaison}\:{a}\:{une}\:{serie}\:{de}\:{riemann} \\ $$$${divergente}..\: \\ $$$$\left.\mathrm{5}\right) \\ $$$${U}_{{n}} =\frac{\left(−\mathrm{1}\right)^{{n}} +{n}}{{n}^{\mathrm{2}} +\mathrm{1}}\:{En}\:{effet}\:,\:\left(−\mathrm{1}\right)^{{n}} +{n}\:\underset{+\infty} {\sim}{n} \\ $$$${de}\:{plus}\:,\:{n}^{\mathrm{2}} +\mathrm{1}\:\underset{+\infty} {\sim}{n}^{\mathrm{2}} \:{ainsi},\:{U}_{{n}} \:\underset{+\infty} {\sim}\frac{\mathrm{1}}{{n}} \\ $$$${et}\:{diverge}\:{par}\:{comparaison}\:{a}\:{une}\:{serie} \\ $$$${de}\:{riemann}\:{divervente}… \\ $$$$\left.\mathrm{6}\right) \\ $$$${U}_{{n}} =\frac{\mathrm{1}}{{n}!}\:{en}\:{fait}\:,\:\forall{n}\geqslant\mathrm{2}\:,\:\frac{\mathrm{1}}{{n}!}\:\leqslant\:\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} }\:{ainsi} \\ $$$${par}\:{comparaison}\:{a}\:{une}\:{serie}\:{geometrique} \\ $$$${convergente}\:,\:{la}\:{serie}\:{converge}.. \\ $$$$\left.\mathrm{7}\right) \\ $$$${U}_{{n}} =\frac{\mathrm{3}^{{n}} +{n}^{\mathrm{4}} }{\mathrm{5}^{{n}} −\mathrm{2}^{{n}} }\:{on}\:{remarque}\:{aisement}\:{que}\: \\ $$$$\mathrm{3}^{{n}} +{n}^{\mathrm{4}} \:\underset{+\infty} {\sim}\mathrm{3}^{{n}} \:{et}\:\mathrm{5}^{{n}} −\mathrm{2}^{{n}} \:\underset{+\infty} {\sim}\mathrm{5}^{{n}} {donc}\:{U}_{{n}} \underset{+\infty} {\sim}\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{{n}} \\ $$$${ainsi},\:{par}\:{comparaison}\:{a}\:{une}\:{serie}\: \\ $$$${geometrique}\:{convergente},\:{la}\:{serie}\: \\ $$$${converge}… \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:……………\mathscr{L}{e}\:{puissant}………….. \\ $$
Commented by Ar Brandon last updated on 20/Nov/21
$$\mathrm{Hum}\:!\:\mathrm{merci}\:\mathrm{bro}.\: \\ $$