■マーダヴァの無限級数(その25)

[1] Πp^2/(p^2−1)=4/3・9/8・25/24・49/48・・・

=Π1/(1−1/p^2)=ζ(2)=π^2/6

=π^2/6

[2] Π(p^2+1)/(p^2−1)=5/3・10/8・26/24・50/48・・・=5/2

ですから,

[3] Πp^2/(p^2+1)=π^2/15=ζ(4)/ζ(2)

が求められます.

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[3]の(証)

  Πp^2/(p^2+1)=Πp^2(p^2−1)/(p^4−1)

 ={Πp^4/(p^4−1)}/{Πp^2/(p^2−1)}

 =ζ(4)/ζ(2)=(π^4/90)/(π^2/6)=π^2/15

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