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Question Number 45080 by turbo msup by abdo last updated on 08/Oct/18

find Σ_(n=0) ^∞  (1/((2n+1)^4 ))

$${find}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{4}} } \\ $$

Commented by maxmathsup by imad last updated on 08/Oct/18

we have Σ_(n=1) ^∞  (1/n^4 ) =Σ_(n=1) ^∞  (1/((2n)^4 )) +Σ_(n=1) ^∞  (1/((2n+1)^4 ))  =(1/(16))Σ_(n=1) ^∞   (1/n^4 ) +Σ_(n=1) ^∞  (1/((2n+1)^4 )) ⇒Σ_(n=1) ^∞  (1/((2n+1)^4 )) =(1−(1/(16)))Σ_(n=1) ^∞  (1/n^4 )  =((15)/(16)). (π^4 /(90)) =((15 π^4 )/(16 .2 .15.3)) =(π^4 /(32.3)) =(π^4 /(96))  ★ Σ_(n=0) ^∞   (1/((2n+1)^4 )) =(π^4 /(96)) ★

$${we}\:{have}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{4}} }\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}\right)^{\mathrm{4}} }\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{4}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{16}}\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}^{\mathrm{4}} }\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{4}} }\:\Rightarrow\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{4}} }\:=\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{16}}\right)\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{4}} } \\ $$$$=\frac{\mathrm{15}}{\mathrm{16}}.\:\frac{\pi^{\mathrm{4}} }{\mathrm{90}}\:=\frac{\mathrm{15}\:\pi^{\mathrm{4}} }{\mathrm{16}\:.\mathrm{2}\:.\mathrm{15}.\mathrm{3}}\:=\frac{\pi^{\mathrm{4}} }{\mathrm{32}.\mathrm{3}}\:=\frac{\pi^{\mathrm{4}} }{\mathrm{96}} \\ $$$$\bigstar\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{4}} }\:=\frac{\pi^{\mathrm{4}} }{\mathrm{96}}\:\bigstar\: \\ $$

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