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Question Number 131428 by pticantor last updated on 04/Feb/21

S=Σ_(n≥1) (6/(n(n+1)(2n+1))) 1) show that S converges 2) give the value of S

$$\:\boldsymbol{{S}}=\underset{\boldsymbol{{n}}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{6}}{\boldsymbol{{n}}\left(\boldsymbol{{n}}+\mathrm{1}\right)\left(\mathrm{2}\boldsymbol{{n}}+\mathrm{1}\right)} \\ $$$$\left.\mathrm{1}\right)\:\boldsymbol{{show}}\:\boldsymbol{{that}}\:\boldsymbol{{S}}\:\boldsymbol{{converges}} \\ $$$$\left.\mathrm{2}\right)\:\boldsymbol{{give}}\:\boldsymbol{{the}}\:\boldsymbol{{value}}\:\boldsymbol{{of}}\:\boldsymbol{{S}} \\ $$

Answered by JDamian last updated on 04/Feb/21

S=Σ_(n≥1) (1/((n(n+1)(2n+1))/6)) (1) a_n =((n(n+1)(2n+1))/6) = 1^2 +2^2 +3^2 +∙∙∙+n^2 (1/a_n ) = (1/(1^2 +2^2 +∙∙∙+n^2 )) <(1/n^2 ) ∀n>1 Σ_(n≥1) (1/a_n ) < Σ_(n≥1) (1/n^2 )=(π^2 /6) S < (π^2 /6)

$${S}=\underset{{n}\geqslant\mathrm{1}} {\Sigma}\:\:\frac{\mathrm{1}}{\frac{\boldsymbol{{n}}\left(\boldsymbol{{n}}+\mathrm{1}\right)\left(\mathrm{2}\boldsymbol{{n}}+\mathrm{1}\right)}{\mathrm{6}}} \\ $$$$\left(\mathrm{1}\right) \\ $$$${a}_{{n}} =\frac{\boldsymbol{{n}}\left(\boldsymbol{{n}}+\mathrm{1}\right)\left(\mathrm{2}\boldsymbol{{n}}+\mathrm{1}\right)}{\mathrm{6}}\:=\:\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\centerdot\centerdot\centerdot+{n}^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{{a}_{{n}} }\:=\:\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\centerdot\centerdot\centerdot+{n}^{\mathrm{2}} }\:<\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:\:\:\:\forall{n}>\mathrm{1} \\ $$$$\underset{{n}\geqslant\mathrm{1}} {\sum}\:\frac{\mathrm{1}}{{a}_{{n}} }\:<\:\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$${S}\:<\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$

Answered by Dwaipayan Shikari last updated on 04/Feb/21

Σ_(n≥1) (6/(n(2n+1)))−Σ_(n≥1) (6/((n+1)(2n+1))) =Σ_(n=0) ^∞ (6/((n+1)(2n+3)))−(Σ_(n≥0) ^∞ (6/((n+1)(2n+1)))−Σ_(n≥0) ^1 (6/((n+1)(2n+1)))) =6(ψ((3/2))+ψ((1/2)))+6 =6(2−4log(2))+6=6(3−4log(2)) ψ((3/2))=2−2log(2) ψ((1/2))=−2log(2)

$$\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{6}}{{n}\left(\mathrm{2}{n}+\mathrm{1}\right)}−\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{6}}{\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{6}}{\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)}−\left(\underset{{n}\geqslant\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{6}}{\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}−\underset{{n}\geqslant\mathrm{0}} {\overset{\mathrm{1}} {\sum}}\frac{\mathrm{6}}{\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}\right) \\ $$$$=\mathrm{6}\left(\psi\left(\frac{\mathrm{3}}{\mathrm{2}}\right)+\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right)+\mathrm{6} \\ $$$$=\mathrm{6}\left(\mathrm{2}−\mathrm{4}{log}\left(\mathrm{2}\right)\right)+\mathrm{6}=\mathrm{6}\left(\mathrm{3}−\mathrm{4}{log}\left(\mathrm{2}\right)\right)\:\:\:\psi\left(\frac{\mathrm{3}}{\mathrm{2}}\right)=\mathrm{2}−\mathrm{2}{log}\left(\mathrm{2}\right)\:\:\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=−\mathrm{2}{log}\left(\mathrm{2}\right) \\ $$$$ \\ $$

Answered by mathmax by abdo last updated on 04/Feb/21

1) we have (6/(n(n+1)(2n+1)))∼(6/(2n^3 ))=(3/n^3 )and Σ (3/n^3 )cv ⇒S cv 2)(S/6)=lim_(n→+∞) w_n with w_n =Σ_(k=1) ^n (1/(k(k+1)(2k+1))) let decompose F(x)=(1/(x(x+1)(2x+1))) ,F(x)=(a/x)+(b/(x+1))+(c/(2x+1)) a=1 ,b =(1/((−1)(−1)))=1 ,c =(1/((−(1/2))(1/2))) =−4 ⇒F(x)=(1/x)+(1/(x+1))−(4/(2x+1)) w_n =Σ_(k^ =1) ^n (1/k)+Σ_(k=1) ^n (1/(k+1))−4 Σ_(k=1) ^n (1/(2k+1)) but Σ_(k=1) ^n (1/k)=H_n ∼log(n)+γ Σ_(k=1) ^n (1/(k+1)) =Σ_(k=2) ^(n+1) (1/k) =H_(n+1) −1 ∼log(n+1)+γ−1 Σ_(k=1) ^n (1/(2k+1)) =(1/3)+(1/5)+...+(1/(2n+1)) =1+(1/2)+(1/3)+....+(1/(2n))+(1/(2n+1)) −1−(1/2)−(1/4)−....−(1/(2n))=H_(2n+1) −(1/2)H_n −1∼log(2n+1)+γ −(1/2)(log(n)+γ)−1 =log(2n+1)−(1/2)log(n)+(γ/2)−1 ⇒ −4Σ_(k=1) ^n (1/(2k+1))∼−4log(2n+1)+2logn−2γ+4 ⇒ W_n ∼log(n)+γ+log(n+1)+γ−1 −4log(2n+1)+2log(n)−2γ +4 ⇒ W_n ∼log(n^3 )+log(n+1)−4log(2n+1)+3 ⇒ W_n ∼log(((n^4 +n^3 )/((2n+1)^4 )))+3 ⇒W_n →3−4log(2) ⇒ S =6 W_∞ =6(3−4log(2))

$$\left.\mathrm{1}\right)\:\mathrm{we}\:\mathrm{have}\:\frac{\mathrm{6}}{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{2n}+\mathrm{1}\right)}\sim\frac{\mathrm{6}}{\mathrm{2n}^{\mathrm{3}} }=\frac{\mathrm{3}}{\mathrm{n}^{\mathrm{3}} }\mathrm{and}\:\Sigma\:\frac{\mathrm{3}}{\mathrm{n}^{\mathrm{3}} }\mathrm{cv}\:\Rightarrow\mathrm{S}\:\mathrm{cv} \\ $$$$\left.\mathrm{2}\right)\frac{\mathrm{S}}{\mathrm{6}}=\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{w}_{\mathrm{n}} \:\mathrm{with}\:\mathrm{w}_{\mathrm{n}} =\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{k}\left(\mathrm{k}+\mathrm{1}\right)\left(\mathrm{2k}+\mathrm{1}\right)} \\ $$$$\mathrm{let}\:\mathrm{decompose}\:\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{x}\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{2x}+\mathrm{1}\right)}\:,\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{a}}{\mathrm{x}}+\frac{\mathrm{b}}{\mathrm{x}+\mathrm{1}}+\frac{\mathrm{c}}{\mathrm{2x}+\mathrm{1}} \\ $$$$\mathrm{a}=\mathrm{1}\:,\mathrm{b}\:=\frac{\mathrm{1}}{\left(−\mathrm{1}\right)\left(−\mathrm{1}\right)}=\mathrm{1}\:,\mathrm{c}\:=\frac{\mathrm{1}}{\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\frac{\mathrm{1}}{\mathrm{2}}}\:=−\mathrm{4}\:\Rightarrow\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{x}+\mathrm{1}}−\frac{\mathrm{4}}{\mathrm{2x}+\mathrm{1}} \\ $$$$\mathrm{w}_{\mathrm{n}} =\sum_{\mathrm{k}^{} =\mathrm{1}} ^{\mathrm{n}} \frac{\mathrm{1}}{\mathrm{k}}+\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{k}+\mathrm{1}}−\mathrm{4}\:\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{2k}+\mathrm{1}}\:\mathrm{but} \\ $$$$\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{k}}=\mathrm{H}_{\mathrm{n}} \sim\mathrm{log}\left(\mathrm{n}\right)+\gamma \\ $$$$\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{k}+\mathrm{1}}\:=\sum_{\mathrm{k}=\mathrm{2}} ^{\mathrm{n}+\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{k}}\:=\mathrm{H}_{\mathrm{n}+\mathrm{1}} −\mathrm{1}\:\sim\mathrm{log}\left(\mathrm{n}+\mathrm{1}\right)+\gamma−\mathrm{1} \\ $$$$\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{2k}+\mathrm{1}}\:=\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}+...+\frac{\mathrm{1}}{\mathrm{2n}+\mathrm{1}}\:=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+....+\frac{\mathrm{1}}{\mathrm{2n}}+\frac{\mathrm{1}}{\mathrm{2n}+\mathrm{1}} \\ $$$$−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}−....−\frac{\mathrm{1}}{\mathrm{2n}}=\mathrm{H}_{\mathrm{2n}+\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}}\mathrm{H}_{\mathrm{n}} −\mathrm{1}\sim\mathrm{log}\left(\mathrm{2n}+\mathrm{1}\right)+\gamma \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{log}\left(\mathrm{n}\right)+\gamma\right)−\mathrm{1}\:=\mathrm{log}\left(\mathrm{2n}+\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\left(\mathrm{n}\right)+\frac{\gamma}{\mathrm{2}}−\mathrm{1}\:\Rightarrow \\ $$$$−\mathrm{4}\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{2k}+\mathrm{1}}\sim−\mathrm{4log}\left(\mathrm{2n}+\mathrm{1}\right)+\mathrm{2logn}−\mathrm{2}\gamma+\mathrm{4}\:\Rightarrow \\ $$$$\mathrm{W}_{\mathrm{n}} \sim\mathrm{log}\left(\mathrm{n}\right)+\gamma+\mathrm{log}\left(\mathrm{n}+\mathrm{1}\right)+\gamma−\mathrm{1}\:−\mathrm{4log}\left(\mathrm{2n}+\mathrm{1}\right)+\mathrm{2log}\left(\mathrm{n}\right)−\mathrm{2}\gamma\:+\mathrm{4}\:\Rightarrow \\ $$$$\mathrm{W}_{\mathrm{n}} \sim\mathrm{log}\left(\mathrm{n}^{\mathrm{3}} \right)+\mathrm{log}\left(\mathrm{n}+\mathrm{1}\right)−\mathrm{4log}\left(\mathrm{2n}+\mathrm{1}\right)+\mathrm{3}\:\Rightarrow \\ $$$$\mathrm{W}_{\mathrm{n}} \sim\mathrm{log}\left(\frac{\mathrm{n}^{\mathrm{4}} \:+\mathrm{n}^{\mathrm{3}} }{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{4}} }\right)+\mathrm{3}\:\Rightarrow\mathrm{W}_{\mathrm{n}} \rightarrow\mathrm{3}−\mathrm{4log}\left(\mathrm{2}\right)\:\Rightarrow \\ $$$$\mathrm{S}\:=\mathrm{6}\:\mathrm{W}_{\infty} =\mathrm{6}\left(\mathrm{3}−\mathrm{4log}\left(\mathrm{2}\right)\right) \\ $$$$ \\ $$

Commented by pticantor last updated on 08/Feb/21

thank you too much

$$\boldsymbol{{thank}}\:\boldsymbol{{you}}\:\boldsymbol{{too}}\:\boldsymbol{{much}} \\ $$

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