Menu Close

Prove-or-disprove-that-S-n-1-1-n-k-k-Z-k-gt-1-is-irrational-




Question Number 3138 by prakash jain last updated on 05/Dec/15
Prove or disprove that  S=Σ_(n=1) ^∞ (1/n^k ) , k∈Z^+ , k>1 is irrational.
ProveordisprovethatS=n=11nk,kZ+,k>1isirrational.
Commented by 123456 last updated on 05/Dec/15
ζ(k)  ζ(2)=(π^2 /6) wich is a rational multiple of  π^2 , so its irational  and in general  ζ(2n)=qπ^(2n)  q∈Q,n∈Z^+  so its irrational  however nothing is know about  ζ(2n+1)  but also  ζ(3)=ζ_3  is irrational (apery constant)
ζ(k)ζ(2)=π26wichisarationalmultipleofπ2,soitsirationalandingeneralζ(2n)=qπ2nqQ,nZ+soitsirrationalhowevernothingisknowaboutζ(2n+1)butalsoζ(3)=ζ3isirrational(aperyconstant)
Commented by Filup last updated on 06/Dec/15
ζ(2n+1)=ζ(2n)(2n+1)
ζ(2n+1)=ζ(2n)(2n+1)
Commented by prakash jain last updated on 06/Dec/15
How did you come up with  ζ(2n+1)=(2n+1)ζ(2n) ?  From what i know read so far  Funtional Equation  ζ(s)=2^s π^(s−1) sin (((πs)/2))Γ(1−s)ζ(1−s)  For s=−2n   ζ(−2n)=0  s=+2n  sin (((πs)/2))Γ(1−s) ≠0 so RHS ≠0
Howdidyoucomeupwithζ(2n+1)=(2n+1)ζ(2n)?FromwhatiknowreadsofarFuntionalEquationζ(s)=2sπs1sin(πs2)Γ(1s)ζ(1s)Fors=2nζ(2n)=0s=+2nsin(πs2)Γ(1s)0soRHS0
Commented by Filup last updated on 06/Dec/15
sorry, my mistake!!!  i was thinking of Γ(2n+1)=Γ(2n)(2n+1)
sorry,mymistake!!!iwasthinkingofΓ(2n+1)=Γ(2n)(2n+1)

Leave a Reply

Your email address will not be published. Required fields are marked *