e^z = ∑{0 ≤ n} z^n/n!,故e^(2z) = ∑{0 ≤ n} (2z)^n/n!= ∑{0 ≤ n} 2^n/n!·z^n,进而得1-e^(2z) = -∑{1 ≤ n} 2^n/n!·z^n.于是(1-e^(2z))/z^2 = -∑{1 ≤ n} 2^n/n!·z^(n-2) = = -∑{-1 ≤ n} 2^(n+2)/(n+2)...