一、f0(z)=∑(k=1…∞)(-1)k-1ln(1+z/k)

=f0(1)-f0(z+1)+ln(z+1).

[f0(1)=ln(π/2)]

二、f1(z)=∑(k=1…∞)(-1)k-1[k·ln(1+z/k)-z]

=f1(1)-f1(z+1)+f0(z)-z.

三、f2(z)=∑(k=1…∞)(-1)k-1[k2·ln(1+z/k)-kz+(1/2)z2]

=f2(1)-f2(z+1)-f0(z)+2f1(z)+z+(1/2)z2.

四、f3(z)=∑(k=1…∞)(-1)k-1[k3·ln(1+z/k)-k2z+(k/2)z2-(1/3)z3].

=f3(1)-f3(z+1)+f0(z)-3f1(z)+3f2(z)-z-(1/2)z2-(1/3)z3.

五、f4(z)=∑(k=1…∞)(-1)k-1[k4·ln(1+z/k)-k3z+(k2/2)z2-(k/3)z3+(1/4)z4].

=f4(1)-f4(z+1)-f0(z)+4f1(z)-6f2(z)+4f3(z)+z+(3/2)z2+z3+(1/4)z4.

六、f5(z)=（略）