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Question Number 72023 by mathmax by abdo last updated on 23/Oct/19

$$find{lim}_{n\to +\mathrm{\infty }}\sum _{1⩽i<j⩽n}\frac{1}{{\left(ij\right)}^2}$$

Commented bymathmax by abdo last updated on 24/Oct/19

$$wehave{\left(\sum _{i=1}^n{x}_i\right)}^2=\sum _{i=1}^n{x}_i^2+2\sum _{1⩽i<j⩽n}{x}_i{x}_j$$ $$let{x}_i=\frac{1}{i^2}\Rightarrow {\left(\sum _{i=1}^n\frac{1}{i^2}\right)}^2=\sum _{i=1}^n\frac{1}{i^4}+2\sum _{1⩽i<j⩽n}\frac{1}{i^2}\times \frac{1}{j^2}$$ $$\Rightarrow {lim}_{n\to +\mathrm{\infty }}{\left(\sum _{i=1}^n\frac{1}{i^2}\right)}^2={lim}_{n\to +\mathrm{\infty }}\sum _{i=1}^n\frac{1}{i^4}+2{lim}_{n\to +\mathrm{\infty }}\sum _{1⩽i<j⩽n}\frac{1}{{\left(ij\right)}^2}\Rightarrow$$ $$2{lim}_{n\to +\mathrm{\infty }}\sum _{1⩽i<j⩽n}\frac{1}{{\left(ij\right)}^2}={\left(\sum _{i=1}^{\mathrm{\infty }}\frac{1}{i^2}\right)}^2-\sum _{n=1}^{\mathrm{\infty }}\frac{1}{i^4}$$ $$={\left(\frac{{\pi }^2}{6}\right)}^2-\xi \left(4\right)=\frac{{\pi }^4}{36}-\xi \left(4\right)\Rightarrow$$ $${lim}_{n\to +\mathrm{\infty }}\sum _{1⩽i<j⩽n}\frac{1}{{\left(ij\right)}^2}=\frac{{\pi }^4}{72}-\frac{\xi \left(4\right)}{2}$$

Answered by mind is power last updated on 23/Oct/19

$$\sum _{1⩽\mathrm{i}<\mathrm{j}⩽\mathrm{n}}=\underset{\mathrm{j}=2}{\sum ^{\mathrm{n}}}\frac{1}{{\mathrm{j}}^2}\underset{\mathrm{i}=1}{\sum ^{\mathrm{j}-1}}\frac{1}{{\mathrm{i}}^2}$$ $$\sum _{1⩽\mathrm{j}<\mathrm{i}⩽\mathrm{n}}\frac{1}{{\left(\mathrm{ij}\right)}^2}=\sum _{1⩽\mathrm{i}<\mathrm{j}⩽\mathrm{n}}\frac{1}{{\left(\mathrm{ij}\right)}^2}$$ $$\underset{\mathrm{j},\mathrm{i}=1}{\sum ^{\mathrm{n}}}\frac{1}{{\left(\mathrm{ij}\right)}^2}=\underset{\mathrm{i}=1}{\sum ^{\mathrm{n}}}\frac{1}{{\mathrm{i}}^2}.\underset{\mathrm{j}=1}{\sum ^{\mathrm{n}}}\frac{1}{{\mathrm{j}}^2}$$ $$\sum _{1⩽\mathrm{j}<\mathrm{i}⩽\mathrm{n}}\frac{1}{{\left(\mathrm{ij}\right)}^2}+\sum _{1⩽\mathrm{i}<\mathrm{j}⩽\mathrm{n}}\frac{1}{{\left(\mathrm{ij}\right)}^2}+\underset{\mathrm{i}=\mathrm{j}=1}{\sum ^{\mathrm{n}}}\frac{1}{{\left(\mathrm{ij}\right)}^2}+\underset{\mathrm{i}=1}{\sum ^{\mathrm{n}}}\frac{1}{{\mathrm{i}}^2}\underset{\mathrm{j}=1}{\sum ^{\mathrm{n}}}\frac{1}{{\mathrm{j}}^2}$$ $$$$ $$2\sum _{1⩽\mathrm{i}<\mathrm{j}⩽\mathrm{n}}\frac{1}{{\left(\mathrm{ij}\right)}^2}=\underset{\mathrm{i}=1}{\sum ^{\mathrm{n}}}\frac{1}{{\mathrm{i}}^2}.\underset{\mathrm{j}.1}{\sum ^{\mathrm{n}}}\frac{1}{{\mathrm{j}}^2}-\underset{1}{\sum ^{\mathrm{n}}}\frac{1}{{\mathrm{j}}^4}$$ $$2\sum _{1⩽\mathrm{i}<\mathrm{j}⩽\mathrm{n}}\frac{1}{{\left(\mathrm{ij}\right)}^2}={\left(\zeta \left(2\right)\right)}^2-\zeta \left(4\right)\Rightarrow$$ $$\sum _{1⩽\mathrm{i}<\mathrm{j}⩽\mathrm{n}}\frac{1}{{\left(\mathrm{ij}\right)}^2}=\frac{\zeta {\left(2\right)}^2-\zeta \left(4\right)}{2}$$

Commented bygunawan last updated on 23/Oct/19

$$\mathrm{you}\mathrm{are}\mathrm{Amazing}$$

Commented bymind is power last updated on 23/Oct/19

$$\mathrm{thanx}\mathrm{sir}$$ $$$$

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