1-n-n-4-

Question Number 147294 by ArielVyny last updated on 19/Jul/21

$$\Sigma\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{4}} }=? \\ $$

Answered by mathmax by abdo last updated on 19/Jul/21

Σ_(n=1) ^∞ (((−1)^n )/n^4 )=(2^(1−4) −1)ξ(4)=((1/8)−1).(π^2 /(90))=−(7/8).(π^2 /(90)) =−((7π^2 )/(720)) remember that Σ(((−1)^n )/n^x ) =(2^(1−x) −1)ξ(x)

$$\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}^{\mathrm{4}} }=\left(\mathrm{2}^{\mathrm{1}−\mathrm{4}} −\mathrm{1}\right)\xi\left(\mathrm{4}\right)=\left(\frac{\mathrm{1}}{\mathrm{8}}−\mathrm{1}\right).\frac{\pi^{\mathrm{2}} }{\mathrm{90}}=−\frac{\mathrm{7}}{\mathrm{8}}.\frac{\pi^{\mathrm{2}} }{\mathrm{90}} \\ $$$$=−\frac{\mathrm{7}\pi^{\mathrm{2}} }{\mathrm{720}}\:\:\mathrm{remember}\:\mathrm{that}\:\Sigma\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}^{\mathrm{x}} }\:=\left(\mathrm{2}^{\mathrm{1}−\mathrm{x}} −\mathrm{1}\right)\xi\left(\mathrm{x}\right) \\ $$

Commented by ArielVyny last updated on 19/Jul/21