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Sachant que pour x>1 :
Γ(x)ζ(x)=∫0+∞et−1tx−1dt
Montrez que pour 1<x<2 :
ζ(x)=x−11+k=1∑+∞(−1)k−1ζ(x+k)Γ(x)Γ(k+2)Γ(x+k)
Observez que la somme du second membre converge pour tout x<2