Suppose that \(f\) is holomorphic in an open set containing the closed unit disc, except for a simple pole at \(z=1\). Let \(f(z)=\sum_{n=1}^{\infty} c_{n} z^{n}\) denote the power series in the open unit disc. Show that \[ \lim _{n \rightarrow \infty} c_{n}=-\lim _{z \rightarrow 1}(z-1) f(z) .\]
Solution
Compute the series expansion of the RHS: \[ (z-1) f(z) &= (z-1) \sum_{n\geq 1} c_n z^k \\ &= -c_1z + \sum_{n\geq 2} (c_{n-1} - c_n) z^n \\ &\convergesto{z\to 1} -c_1 + \sum_{n\geq 2} c_{n-1} - c_n \\ &\da \lim_{N\to\infty} -c_1 z + \sum_{n=2}^N c_{n-1} - c_n \\ &= \lim_{N\to\infty} -c_N ,\] where we’ve used that the sum is telescoping.