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Question Number 58354 by maxmathsup by imad last updated on 21/Apr/19
$${let}\:{U}_{{n}} =\frac{\mathrm{1}^{\mathrm{2}} \:+\mathrm{2}^{\mathrm{2}} \:+\mathrm{3}^{\mathrm{2}} \:+....+{n}^{\mathrm{2}} }{\mathrm{1}^{\mathrm{4}} \:+\mathrm{2}^{\mathrm{4}} \:+\mathrm{3}^{\mathrm{4}} \:+....+{n}^{\mathrm{4}} } \\ $$$$\left.\mathrm{1}\right){find}\:{lim}_{{n}\rightarrow+\infty} {U}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:{U}_{{n}} \\ $$
Answered by tanmay last updated on 22/Apr/19
$${U}_{{n}} \:\:{here}\:=\frac{{N}_{{r}} }{{D}_{{r}} } \\ $$$${N}_{{r}} =\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{r}^{\mathrm{2}} =\frac{{n}^{\mathrm{3}} }{\mathrm{3}}+\frac{{n}^{\mathrm{2}} }{\mathrm{2}}+\frac{{n}}{\mathrm{6}} \\ $$$${D}_{{r}} =\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{r}^{\mathrm{4}} =\frac{{n}^{\mathrm{5}} }{\mathrm{5}}+\frac{{n}^{\mathrm{4}} }{\mathrm{2}}+\frac{{n}^{\mathrm{3}} }{\mathrm{3}}−\frac{{n}}{\mathrm{30}} \\ $$$${each}\:{term}\:{of}\:{N}_{{r}} <\:{each}\:{term}\:{of}\:{D}_{{r}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\frac{{n}^{\mathrm{3}} }{\mathrm{3}}+\frac{{n}^{\mathrm{2}} }{\mathrm{2}}+\frac{{n}}{\mathrm{6}}}{\frac{{n}^{\mathrm{5}} }{\mathrm{5}}+\frac{{n}^{\mathrm{4}} }{\mathrm{2}}+\frac{{n}^{\mathrm{3}} }{\mathrm{3}}−\frac{{n}}{\mathrm{30}}} \\ $$$${devide}\:{N}_{{r}} \:{and}\:{D}_{{r}} \:{by}\:\:{n}^{\mathrm{5}} \rightarrow \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\frac{\mathrm{1}}{\mathrm{3}{n}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{6}{n}^{\mathrm{4}} }}{\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{2}{n}}+\frac{\mathrm{1}}{\mathrm{3}{n}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{30}{n}^{\mathrm{4}} }} \\ $$$$=\frac{\mathrm{0}+\mathrm{0}+\mathrm{0}}{\frac{\mathrm{1}}{\mathrm{5}}+\mathrm{0}+\mathrm{0}−\mathrm{0}} \\ $$$$=\mathrm{0} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$