Question Number 26751 by abdo imad last updated on 28/Dec/17
$${by}\:{using}\:{fourier}\:{serie}\:{find}\:{the}\:{value}\:{of}\:\sum_{{n}=\mathrm{0}} ^{\propto} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$
Commented by abdo imad last updated on 30/Dec/17
$${we}\:{know}\:{that}\:/{x}/=\:\frac{\pi}{\mathrm{2}}\:−\frac{\mathrm{4}}{\pi}\:\sum_{{p}=\mathrm{0}} ^{\propto} \:\frac{{cos}\left(\mathrm{2}{p}+\mathrm{1}\right){x}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\left({the}\:{fuction}\:{is}\:\mathrm{2}\pi\:{periodic}\:{even}\right)\Rightarrow{let}\:{take}\:{x}=\mathrm{0} \\ $$$$\mathrm{0}=\:\frac{\pi}{\mathrm{2}}\:\:−\frac{\mathrm{4}}{\pi}\:\sum_{{p}=\mathrm{0}} ^{\propto} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow\:\:\sum_{{p}=\mathrm{0}} ^{\propto} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} }\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:\:. \\ $$