Expansion for a reciprocal

One way: polynomial long division. \[ {1\over e^z-1} &= {1\over z + {1\over 2}z^2 + {1\over 6}z^2 + \cdots } \\ &= {1\over z}\qty{1 - {1\over 2}z + \qty{-{1\over 6} + {1\over 4} }z^2 + \cdots } \\ &= \inverseof{z} - {1\over 2} + {1\over 12}z + \bigo(z^3) .\] Alternatively, use geometric series. Note that something like \({1\over 1-e^z} = \sum_{k\geq 0} e^{kz}\) won’t converge, and won’t even be calculable since each \(e^{kz}\) contributes a constant term! \[ {1\over e^z-1} &= {1\over z + {1\over 2}z^2 + {1\over 6}z^3 + \cdots } \\ &= {1\over z(1 + {1\over 2}z + {1\over 6}z^2 + \cdots) } \\ &= \inverseof{z} {1\over 1 + q(z) } \qquad q(z) \da {1\over 2}z + {1\over 6}z^2 + \cdots \\ &= \inverseof{z} \sum_{k\geq 0}(-q(z))^k \\ &= \inverseof{z} \qty{1 - q(z) + q(z)^2 - \cdots } \\ &= \inverseof{z}\qty{1 - \qty{{1\over 2}z + {1\over 6}z^2 + \cdots } + \qty{{1\over 2}z + {1\over 6}z^2 + \cdots }^2 - \cdots } \\ &= \inverseof{z}\qty{ 1 - {1\over 2} z + \qty{-{1\over 6} + \qty{1\over 2}^2 } z^2 + \bigo(z^3) } \\ &= \inverseof{z} - {1\over 2} + {1\over 12}z + \bigo(z^2) .\]