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Question Number 60504 by prof Abdo imad last updated on 21/May/19

let S_n =Σ_(k=1) ^n ((1^2 +2^2 +...k^2 )/(1^4 +2^4 +...+k^4 )) study the convergence of S_n

$$\:{let}\:{S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}^{\mathrm{2}} \:+\mathrm{2}^{\mathrm{2}} \:+...{k}^{\mathrm{2}} }{\mathrm{1}^{\mathrm{4}} \:+\mathrm{2}^{\mathrm{4}} \:+...+{k}^{\mathrm{4}} } \\ $$$${study}\:{the}\:{convergence}\:{of}\:{S}_{{n}} \\ $$

Commented by maxmathsup by imad last updated on 22/May/19

we have 1^2 +2^2 +3^2 +...+k^2 =((k(k+1)(2k+1))/6) 1^4 +2^4 +3^4 +...+k^4 =((k(k+1)(2k+1)(3k^2 +3k−1))/(30)) ⇒ S_n = Σ_(k=1) ^n (5/(3k^2 +3k−1)) =Σ_(k=1) ^n U_k ⇒lim_(n→+∞) S_n =Σ_(n=1) ^∞ (5/(3n^2 +3n −1)) =S U_n = (5/(n^2 (3 +(3/n)−(1/n^2 )))) ∼ (5/n^2 ) but Σ (5/n^2 ) converges ⇒ S converges

$${we}\:{have}\:\mathrm{1}^{\mathrm{2}} \:+\mathrm{2}^{\mathrm{2}} \:+\mathrm{3}^{\mathrm{2}} \:+...+{k}^{\mathrm{2}} =\frac{{k}\left({k}+\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$$\mathrm{1}^{\mathrm{4}} +\mathrm{2}^{\mathrm{4}} \:+\mathrm{3}^{\mathrm{4}} \:+...+{k}^{\mathrm{4}} \:=\frac{{k}\left({k}+\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{1}\right)\left(\mathrm{3}{k}^{\mathrm{2}} \:+\mathrm{3}{k}−\mathrm{1}\right)}{\mathrm{30}}\:\Rightarrow \\ $$$${S}_{{n}} =\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{5}}{\mathrm{3}{k}^{\mathrm{2}} \:+\mathrm{3}{k}−\mathrm{1}}\:=\sum_{{k}=\mathrm{1}} ^{{n}} \:{U}_{{k}} \:\Rightarrow{lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} =\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{5}}{\mathrm{3}{n}^{\mathrm{2}} \:+\mathrm{3}{n}\:−\mathrm{1}}\:={S} \\ $$$${U}_{{n}} =\:\frac{\mathrm{5}}{{n}^{\mathrm{2}} \left(\mathrm{3}\:+\frac{\mathrm{3}}{{n}}−\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)}\:\sim\:\frac{\mathrm{5}}{{n}^{\mathrm{2}} }\:\:\:{but}\:\Sigma\:\frac{\mathrm{5}}{{n}^{\mathrm{2}} }\:{converges}\:\Rightarrow\:{S}\:{converges} \\ $$

Commented by maxmathsup by imad last updated on 22/May/19

U_n ∼ (5/(3n^2 )) (n→+∞)

$${U}_{{n}} \sim\:\frac{\mathrm{5}}{\mathrm{3}{n}^{\mathrm{2}} }\:\left({n}\rightarrow+\infty\right) \\ $$

Answered by Prithwish sen last updated on 22/May/19

a_n =((30n(n+1)(2n+1))/(6n(n+1)(2n+1)(3n^2 +3n−1))) =(5/(3n^2 +3n−1)) ∴ By Raabe′s test Lt_( n→∞) R_n = lt_(n→∞ ) n((a_n /a_(n+1) ) −1) =Lt_(n→∞) n(((3n^2 +9n+5)/(3n^2 +3n−1)) −1) =Lt_(n→∞) (((6+(6/n))/(3+(3/n)−(1/n^2 )))) = (6/3) =2>1 ∴ According to Raabe′s test S_n converges Is it ok? Please share your feedbacks

$$\mathrm{a}_{\mathrm{n}} =\frac{\mathrm{30n}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{2n}+\mathrm{1}\right)}{\mathrm{6n}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{2n}+\mathrm{1}\right)\left(\mathrm{3n}^{\mathrm{2}} +\mathrm{3n}−\mathrm{1}\right)} \\ $$$$\:\:\:\:=\frac{\mathrm{5}}{\mathrm{3n}^{\mathrm{2}} +\mathrm{3n}−\mathrm{1}} \\ $$$$\therefore\:\mathrm{By}\:\mathrm{Raabe}'\mathrm{s}\:\mathrm{test} \\ $$$$\mathrm{Lt}_{\:\mathrm{n}\rightarrow\infty} \mathrm{R}_{\mathrm{n}} =\:\mathrm{lt}_{\mathrm{n}\rightarrow\infty\:\:\:} \mathrm{n}\left(\frac{\mathrm{a}_{\mathrm{n}} }{\mathrm{a}_{\mathrm{n}+\mathrm{1}} }\:−\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{Lt}_{\mathrm{n}\rightarrow\infty} \:\mathrm{n}\left(\frac{\mathrm{3n}^{\mathrm{2}} +\mathrm{9n}+\mathrm{5}}{\mathrm{3n}^{\mathrm{2}} +\mathrm{3n}−\mathrm{1}}\:−\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{Lt}_{\mathrm{n}\rightarrow\infty} \left(\frac{\mathrm{6}+\frac{\mathrm{6}}{\mathrm{n}}}{\mathrm{3}+\frac{\mathrm{3}}{\mathrm{n}}−\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{6}}{\mathrm{3}}\:=\mathrm{2}>\mathrm{1} \\ $$$$\therefore\:\mathrm{According}\:\mathrm{to}\:\mathrm{Raabe}'\mathrm{s}\:\mathrm{test}\:\mathrm{S}_{\mathrm{n}} \:\mathrm{converges} \\ $$$$\mathrm{Is}\:\mathrm{it}\:\mathrm{ok}?\:\mathrm{Please}\:\mathrm{share}\:\mathrm{your}\:\mathrm{feedbacks} \\ $$$$ \\ $$

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