Question Number 52682 by maxmathsup by imad last updated on 11/Jan/19
$${find}\:{nature}\:{of}\:{the}\:{serie}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\sqrt{{n}+\mathrm{1}}−\sqrt{{n}}}{{nln}\left({n}+\mathrm{1}\right)} \\ $$
Commented by Abdo msup. last updated on 12/Jan/19
$${let}\:{u}_{{n}} =\frac{\sqrt{{n}+\mathrm{1}}−\sqrt{{n}}}{{nln}\left({n}+\mathrm{1}\right)}\:\:{we}\:{have} \\ $$$${u}_{{n}} =\frac{\mathrm{1}}{\left(\sqrt{{n}+\mathrm{1}}+\sqrt{{n}}\right){nln}\left({n}+\mathrm{1}\right)} \\ $$$$\sqrt{{n}+\mathrm{1}}+\sqrt{{n}}\:\sim\mathrm{2}\sqrt{{n}}\:{and}\:{ln}\left({n}+\mathrm{1}\right)={ln}\left({n}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)\right) \\ $$$${ln}\left({n}\right)\:+{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)\:\sim{ln}\left({n}\right)+\frac{\mathrm{1}}{{n}}\:\Rightarrow \\ $$$${u}_{{n}} \sim\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{{n}}\left({ln}\left({n}\right)+\frac{\mathrm{1}}{{n}}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{{n}}{ln}\left({n}\right)\:+\frac{\mathrm{2}}{\:\sqrt{{n}}}}\:\sim\frac{\mathrm{1}}{\mathrm{2}\sqrt{{n}}{ln}\left({n}\right)} \\ $$$${the}\:{sequence}\:{v}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}\sqrt{{n}}{ln}\left({n}\right)}\:{is}\:{decreasing}\:\Rightarrow \\ $$$$\Sigma{v}_{{n}} \:{and}\:\:\int_{\mathrm{2}} ^{+\infty} \:\:\:\frac{{dx}}{\mathrm{2}\sqrt{{x}}{ln}\left({x}\right)}\:{have}\:{the}\:{same}\:{nature}\:{of} \\ $$$${convergence}\:{but}\: \\ $$$$\int_{\mathrm{2}} ^{+\infty} \:\:\frac{{dx}}{\mathrm{2}\sqrt{{x}}{ln}\left({x}\right)}\:=_{{ln}\left({x}\right)={t}} \:\:\:\:\int_{{ln}\left(\mathrm{2}\right)} ^{+\infty} \:\:\:\frac{{e}^{{t}} {dt}}{\mathrm{2}{e}^{\frac{{t}}{\mathrm{2}}} {t}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{{ln}\left(\mathrm{2}\right)} ^{+\infty} \:\:\:\:\frac{{e}^{\frac{{t}}{\mathrm{2}}} }{{t}}\:{dt}\:\:{and}\:{this}\:{integral}\:{diverges} \\ $$$$\left({lim}_{{t}\rightarrow+\infty} \:{t}\:.\frac{{e}^{\frac{{t}}{\mathrm{2}}} }{{t}}\:=+\infty+\:{so}\:\Sigma\:{u}_{{n}} \:{diverges}.\right. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 12/Jan/19
$$\frac{{T}_{{n}} }{{T}_{{n}−\mathrm{1}} }=\frac{\sqrt{{n}+\mathrm{1}}\:−\sqrt{{n}}}{{nln}\left({n}+\mathrm{1}\right)}×\frac{\left({n}−\mathrm{1}\right){ln}\left({n}\right)}{\:\sqrt{{n}}\:−\sqrt{{n}−\mathrm{1}}} \\ $$$$ \\ $$$$=\frac{\mathrm{1}−\sqrt{\frac{{n}}{{n}+\mathrm{1}}}}{\:\sqrt{\frac{{n}}{{n}+\mathrm{1}}}\:−\sqrt{\frac{{n}−\mathrm{1}}{{n}+\mathrm{1}}}}×\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)×\frac{{lnn}}{{ln}\left({n}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}−\sqrt{\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{{n}}}}}{\:\sqrt{\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{{n}}}}}×\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)×\frac{{lnn}}{{ln}\left({n}+\mathrm{1}\right)} \\ $$$${when}\:{n}\rightarrow\infty \\ $$$$=\frac{\mathrm{1}−\sqrt{\frac{\mathrm{1}}{\mathrm{1}+\mathrm{0}}}}{\:\sqrt{\frac{\mathrm{1}}{\mathrm{1}+\mathrm{0}}}}×\left(\mathrm{1}−\mathrm{0}\right)×\frac{{lnn}}{{ln}\left({n}+\mathrm{1}\right)} \\ $$$$=\mathrm{0}\:\:{so}\:{convergent}… \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{li}{p}}\frac{{T}_{{n}} }{{T}_{{n}−\mathrm{1}} } \\ $$$$ \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 12/Jan/19
