Question Number 45080 by turbo msup by abdo last updated on 08/Oct/18
$${find}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{4}} } \\ $$
Commented by maxmathsup by imad last updated on 08/Oct/18
$${we}\:{have}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{4}} }\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}\right)^{\mathrm{4}} }\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{4}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{16}}\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}^{\mathrm{4}} }\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{4}} }\:\Rightarrow\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{4}} }\:=\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{16}}\right)\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{4}} } \\ $$$$=\frac{\mathrm{15}}{\mathrm{16}}.\:\frac{\pi^{\mathrm{4}} }{\mathrm{90}}\:=\frac{\mathrm{15}\:\pi^{\mathrm{4}} }{\mathrm{16}\:.\mathrm{2}\:.\mathrm{15}.\mathrm{3}}\:=\frac{\pi^{\mathrm{4}} }{\mathrm{32}.\mathrm{3}}\:=\frac{\pi^{\mathrm{4}} }{\mathrm{96}} \\ $$$$\bigstar\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{4}} }\:=\frac{\pi^{\mathrm{4}} }{\mathrm{96}}\:\bigstar\: \\ $$