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Question Number 72023 by mathmax by abdo last updated on 23/Oct/19
find lim_(n→+∞) Σ_(1≤i<j≤n) (1/((ij)^2 ))
$${find}\:{lim}_{{n}\rightarrow+\infty} \:\:\:\sum_{\mathrm{1}\leqslant{i}<{j}\leqslant{n}} \:\:\frac{\mathrm{1}}{\left({ij}\right)^{\mathrm{2}} } \\ $$
Commented by mathmax by abdo last updated on 24/Oct/19
wehave (Σ_(i=1) ^n x_i )^2 =Σ_(i=1) ^n x_i ^2 +2Σ_(1≤i<j≤n) x_i x_j let x_i =(1/i^2 ) ⇒(Σ_(i=1) ^n (1/i^2 ))^2 =Σ_(i=1) ^n (1/i^4 ) +2Σ_(1≤i<j≤n) (1/i^2 )×(1/j^2 ) ⇒lim_(n→+∞) (Σ_(i=1) ^n (1/i^2 ))^2 =lim_(n→+∞) Σ_(i=1) ^n (1/i^4 ) +2lim_(n→+∞) Σ_(1≤i<j≤n) (1/((ij)^2 )) ⇒ 2lim_(n→+∞) Σ_(1≤i<j≤n) (1/((ij)^2 )) =(Σ_(i=1) ^∞ (1/i^2 ))^2 −Σ_(n=1) ^∞ (1/i^4 ) =((π^2 /6))^2 −ξ(4) =(π^4 /(36)) −ξ(4) ⇒ lim_(n→+∞) Σ_(1≤i<j≤n) (1/((ij)^2 )) =(π^4 /(72)) −((ξ(4))/2)
$${wehave}\:\left(\sum_{{i}=\mathrm{1}} ^{{n}} {x}_{{i}} \right)^{\mathrm{2}} =\sum_{{i}=\mathrm{1}} ^{{n}} \:{x}_{{i}} ^{\mathrm{2}} \:+\mathrm{2}\sum_{\mathrm{1}\leqslant{i}<{j}\leqslant{n}} \:\:{x}_{{i}} {x}_{{j}} \\ $$$${let}\:{x}_{{i}} =\frac{\mathrm{1}}{{i}^{\mathrm{2}} }\:\Rightarrow\left(\sum_{{i}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{i}^{\mathrm{2}} }\right)^{\mathrm{2}} =\sum_{{i}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{i}^{\mathrm{4}} }\:+\mathrm{2}\sum_{\mathrm{1}\leqslant{i}<{j}\leqslant{n}} \:\:\frac{\mathrm{1}}{{i}^{\mathrm{2}} }×\frac{\mathrm{1}}{{j}^{\mathrm{2}} } \\ $$$$\Rightarrow{lim}_{{n}\rightarrow+\infty} \left(\sum_{{i}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{i}^{\mathrm{2}} }\right)^{\mathrm{2}} ={lim}_{{n}\rightarrow+\infty} \sum_{{i}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{i}^{\mathrm{4}} }\:+\mathrm{2}{lim}_{{n}\rightarrow+\infty} \sum_{\mathrm{1}\leqslant{i}<{j}\leqslant{n}} \frac{\mathrm{1}}{\left({ij}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\mathrm{2}{lim}_{{n}\rightarrow+\infty} \:\sum_{\mathrm{1}\leqslant{i}<{j}\leqslant{n}} \:\:\frac{\mathrm{1}}{\left({ij}\right)^{\mathrm{2}} }\:\:=\left(\sum_{{i}=\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{{i}^{\mathrm{2}} }\right)^{\mathrm{2}} −\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{i}^{\mathrm{4}} } \\ $$$$=\left(\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\right)^{\mathrm{2}} −\xi\left(\mathrm{4}\right)\:=\frac{\pi^{\mathrm{4}} }{\mathrm{36}}\:−\xi\left(\mathrm{4}\right)\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} \sum_{\mathrm{1}\leqslant{i}<{j}\leqslant{n}} \:\:\frac{\mathrm{1}}{\left({ij}\right)^{\mathrm{2}} }\:=\frac{\pi^{\mathrm{4}} }{\mathrm{72}}\:−\frac{\xi\left(\mathrm{4}\right)}{\mathrm{2}} \\ $$
Answered by mind is power last updated on 23/Oct/19
Σ_(1≤i<j≤n) =Σ_(j=2) ^n (1/j^2 )Σ_(i=1) ^(j−1) (1/i^2 ) Σ_(1≤j<i≤n) (1/((ij)^2 ))=Σ_(1≤i<j≤n) (1/((ij)^2 )) Σ_(j,i=1) ^n (1/((ij)^2 ))=Σ_(i=1) ^n (1/i^2 ).Σ_(j=1) ^n (1/j^2 ) Σ_(1≤j<i≤n) (1/((ij)^2 ))+Σ_(1≤i<j≤n) (1/((ij)^2 ))+Σ_(i=j=1) ^n (1/((ij)^2 ))+Σ_(i=1) ^n (1/i^2 )Σ_(j=1) ^n (1/j^2 ) 2Σ_(1≤i<j≤n) (1/((ij)^2 ))=Σ_(i=1) ^n (1/i^2 ).Σ_(j.1) ^n (1/j^2 )−Σ_1 ^n (1/j^4 ) 2Σ_(1≤i<j≤n) (1/((ij)^2 ))=(ζ(2))^2 −ζ(4)⇒ Σ_(1≤i<j≤n) (1/((ij)^2 ))=((ζ(2)^2 −ζ(4))/2)
$$\sum_{\mathrm{1}\leqslant\mathrm{i}<\mathrm{j}\leqslant\mathrm{n}} =\underset{\mathrm{j}=\mathrm{2}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{j}^{\mathrm{2}} }\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{j}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\mathrm{i}^{\mathrm{2}} } \\ $$$$\sum_{\mathrm{1}\leqslant\mathrm{j}<\mathrm{i}\leqslant\mathrm{n}} \frac{\mathrm{1}}{\left(\mathrm{ij}\right)^{\mathrm{2}} }=\sum_{\mathrm{1}\leqslant\mathrm{i}<\mathrm{j}\leqslant\mathrm{n}} \frac{\mathrm{1}}{\left(\mathrm{ij}\right)^{\mathrm{2}} } \\ $$$$\underset{\mathrm{j},\mathrm{i}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{ij}\right)^{\mathrm{2}} }=\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{i}^{\mathrm{2}} }.\underset{\mathrm{j}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{j}^{\mathrm{2}} } \\ $$$$\sum_{\mathrm{1}\leqslant\mathrm{j}<\mathrm{i}\leqslant\mathrm{n}} \frac{\mathrm{1}}{\left(\mathrm{ij}\right)^{\mathrm{2}} }+\sum_{\mathrm{1}\leqslant\mathrm{i}<\mathrm{j}\leqslant\mathrm{n}} \frac{\mathrm{1}}{\left(\mathrm{ij}\right)^{\mathrm{2}} }+\underset{\mathrm{i}=\mathrm{j}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{ij}\right)^{\mathrm{2}} }+\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{i}^{\mathrm{2}} }\underset{\mathrm{j}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{j}^{\mathrm{2}} } \\ $$$$ \\ $$$$\mathrm{2}\underset{\mathrm{1}\leqslant\mathrm{i}<\mathrm{j}\leqslant\mathrm{n}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{ij}\right)^{\mathrm{2}} }=\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{i}^{\mathrm{2}} }.\underset{\mathrm{j}.\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{j}^{\mathrm{2}} }−\underset{\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{j}^{\mathrm{4}} } \\ $$$$\mathrm{2}\underset{\mathrm{1}\leqslant\mathrm{i}<\mathrm{j}\leqslant\mathrm{n}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{ij}\right)^{\mathrm{2}} }=\left(\zeta\left(\mathrm{2}\right)\right)^{\mathrm{2}} −\zeta\left(\mathrm{4}\right)\Rightarrow \\ $$$$\underset{\mathrm{1}\leqslant\mathrm{i}<\mathrm{j}\leqslant\mathrm{n}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{ij}\right)^{\mathrm{2}} }=\frac{\zeta\left(\mathrm{2}\right)^{\mathrm{2}} −\zeta\left(\mathrm{4}\right)}{\mathrm{2}} \\ $$
Commented by gunawan last updated on 23/Oct/19
you are Amazing
$$\mathrm{you}\:\mathrm{are}\:\mathrm{Amazing} \\ $$
Commented by mind is power last updated on 23/Oct/19
thanx sir
$$\mathrm{thanx}\:\mathrm{sir} \\ $$$$ \\ $$

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