Question Number 26749 by abdo imad last updated on 28/Dec/17
$${let}\:{give}\:\:{S}_{{n}} \:=\:\:\sum_{\mathrm{1}\leqslant{i}<{j}\leqslant{n}} \:\:\:\frac{\mathrm{1}}{{i}^{\mathrm{2}} {j}^{\mathrm{2}} }\:\:\:{find}\:{lim}_{{n}−>\propto} \:\:{S}_{{n}} \:\:. \\ $$
Commented byabdo imad last updated on 02/Jan/18
$${we}\:{have}\:\left(\:\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:\right)^{\mathrm{2}} =\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}^{\mathrm{4}} }\:\:+\:\mathrm{2}\:\sum_{\mathrm{1}\leqslant{i}<{j}\leqslant{n}} \:\:\frac{\mathrm{1}}{{i}^{\mathrm{2}\:} {j}^{\mathrm{2}} } \\ $$ $${S}_{{n}} =\:\frac{\mathrm{1}}{\mathrm{2}}\left(\left(\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\right)^{\mathrm{2}} −\:\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\frac{\mathrm{1}}{{k}^{\mathrm{4}} }\:\right) \\ $$ $$\Rightarrow{lim}_{{n}−>\propto} \:{S}_{{n}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\:\:\left(\xi\left(\mathrm{2}\right)\right)^{\mathrm{2}} −\xi\left(\mathrm{4}\right)\right)\:\:{we}\:{know}\:{that}\:\xi\left(\mathrm{2}\right)=\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$ $${and}\:\:\xi\left(\mathrm{4}\right)\:=\:\:\frac{\pi^{\mathrm{4}} }{\mathrm{90}} \\ $$ $$\Rightarrow\:{lim}_{{n}−>\propto} \:{S}_{{n}} \:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\:\:\frac{\pi^{\mathrm{4}} }{\mathrm{36}}\:−\:\frac{\pi^{\mathrm{4}} }{\mathrm{90}}\:\:\right). \\ $$