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Question Number 26749 by abdo imad last updated on 28/Dec/17

$$letgive{S}_n=\sum _{1⩽i<j⩽n}\frac{1}{i^2{j}^2}find{lim}_{n->\propto }{S}_n.$$

Commented byabdo imad last updated on 02/Jan/18

$$wehave{\left(\sum _{k=1}^{k=n}\frac{1}{k^2}\right)}^2=\sum _{k=1}^n\frac{1}{k^4}+2\sum _{1⩽i<j⩽n}\frac{1}{i^2{j}^2}$$ $$S_n=\frac{1}{2}({\left(\sum _{k=1}^n\frac{1}{k^2}\right)}^2-\sum _{k=1}^{k=n}\frac{1}{k^4})$$ $$\Rightarrow {lim}_{n->\propto }{S}_n=\frac{1}{2}({\left(\xi \left(2\right)\right)}^2-\xi \left(4\right))weknowthat\xi \left(2\right)=\frac{{\pi }^2}{6}$$ $$and\xi \left(4\right)=\frac{{\pi }^4}{90}$$ $$\Rightarrow {lim}_{n->\propto }{S}_n=\frac{1}{2}(\frac{{\pi }^4}{36}-\frac{{\pi }^4}{90}).$$

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