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Question Number 42493 by maxmathsup by imad last updated on 26/Aug/18

calculate lim_(n→+∞) Σ_(1≤i<j≤n) (1/(i^x j^x )) with x>1 for that consider ξ(x) =Σ_(n=1) ^∞ (1/n^x ) 2) calculate lim_(n→+∞) Σ_(1≤i<j≤n) (1/((ij)^2 )) .

$$\:{calculate}\:{lim}_{{n}\rightarrow+\infty} \:\:\:\sum_{\mathrm{1}\leqslant{i}<{j}\leqslant{n}} \:\:\:\:\:\frac{\mathrm{1}}{{i}^{{x}} {j}^{{x}} }\:\:\:{with}\:\:{x}>\mathrm{1}\:\:{for}\:{that}\:{consider} \\ $$ $$\xi\left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}^{{x}} } \\ $$ $$\left.\mathrm{2}\right)\:{calculate}\:{lim}_{{n}\rightarrow+\infty} \:\sum_{\mathrm{1}\leqslant{i}<{j}\leqslant{n}} \:\:\:\:\:\:\frac{\mathrm{1}}{\left({ij}\right)^{\mathrm{2}} }\:. \\ $$

Commented bymaxmathsup by imad last updated on 29/Aug/18

let S_n (x)= Σ_(1≤i<j≤n) (1/(i^x j^x )) and ξ_n (x)=Σ_(k=1) ^n (1/k^x ) we have ( Σ_(k=1) ^n (1/k^x ))^2 =Σ_(k=1) ^n (1/k^(2x) ) +2 Σ_(1≤i<j≤n) (1/(i^x j^x )) let passe to limit(n→+∞) we get (ξ(x))^2 = ξ(2x) + 2 lim_(n→+∞) S_n (x) ⇒ lim_(n→+∞) S_n (x) = (1/2){ (ξ(x))^2 −ξ(2x)} 2) we have lim_(n→+∞) Σ_(1≤i<j≤n) (1/(i^2 j^2 )) =lim_(n→+∞) S_n (2) =(1/2) { (ξ(2))^2 −ξ(4)} =(1/2){ ((π^2 /6))^2 −ξ(4)} =(1/2){ (π^4 /(36)) −ξ(4)} .

$${let}\:\:{S}_{{n}} \left({x}\right)=\:\sum_{\mathrm{1}\leqslant{i}<{j}\leqslant{n}} \:\:\:\:\:\frac{\mathrm{1}}{{i}^{{x}} \:{j}^{{x}} }\:\:\:\:\:{and}\:\xi_{{n}} \left({x}\right)=\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}^{{x}} }\:\:\:{we}\:{have}\: \\ $$ $$\left(\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\:\frac{\mathrm{1}}{{k}^{{x}} }\right)^{\mathrm{2}} \:\:=\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\:\frac{\mathrm{1}}{{k}^{\mathrm{2}{x}} }\:+\mathrm{2}\:\sum_{\mathrm{1}\leqslant{i}<{j}\leqslant{n}} \:\:\:\:\:\:\frac{\mathrm{1}}{{i}^{{x}} \:{j}^{{x}} }\:\:\:{let}\:{passe}\:{to}\:{limit}\left({n}\rightarrow+\infty\right) \\ $$ $${we}\:{get}\:\:\left(\xi\left({x}\right)\right)^{\mathrm{2}} \:\:\:=\:\xi\left(\mathrm{2}{x}\right)\:+\:\mathrm{2}\:{lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} \left({x}\right)\:\Rightarrow \\ $$ $${lim}_{{n}\rightarrow+\infty} \:\:\:{S}_{{n}} \left({x}\right)\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\:\left(\xi\left({x}\right)\right)^{\mathrm{2}} \:−\xi\left(\mathrm{2}{x}\right)\right\} \\ $$ $$\left.\mathrm{2}\right)\:{we}\:{have}\:{lim}_{{n}\rightarrow+\infty} \:\:\sum_{\mathrm{1}\leqslant{i}<{j}\leqslant{n}} \:\:\frac{\mathrm{1}}{{i}^{\mathrm{2}} {j}^{\mathrm{2}} }\:={lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} \left(\mathrm{2}\right) \\ $$ $$=\frac{\mathrm{1}}{\mathrm{2}}\:\left\{\:\left(\xi\left(\mathrm{2}\right)\right)^{\mathrm{2}} \:−\xi\left(\mathrm{4}\right)\right\}\:=\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\:\left(\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\right)^{\mathrm{2}} \:−\xi\left(\mathrm{4}\right)\right\} \\ $$ $$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\frac{\pi^{\mathrm{4}} }{\mathrm{36}}\:−\xi\left(\mathrm{4}\right)\right\}\:. \\ $$

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