■ベキ和の公式の整除性(その19)

  z・exp(zx)/(exp(z)−1)

=(1+xz+1/2!x^2z^2+1/3!x^3z^3+・・・)(B0+B1/1!z+B2/2!z^2+B3/3!z^3+・・・)

  z・exp(z(x+1)/(exp(z)−1)

=z・exp(zx)・exp(z)/(exp(z)−1)

=z・exp(zx)/(exp(z)−1)+z・exp(zx)

 z^sの項を比較すると

  Bs(x+1)−Bs(x)=sx^s-1

  Bs+1(x+1)−Bs+1(x)=(s+1)x^s

x=0,1,2,・・・,nとして加えると

  Σk^s=1^s+2^s+3^s+・・・+n^s

=1/(s+1)・{Bs+1(n)−Bs+1(0)}+n^s

=n^s+1/(s+1)+n^s/2+(p,1)B1/2・n^s-1+(p,3)B4/4・n^s-3+・・・

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