Let \(R>1\) and let \(f\) be analytic on \(|z|<R\) except for a simple pole at \(z=1\). Suppose \[ f(z)=\sum_{n=0}^\infty a_nz^n \qquad(|z|<1). \] Show that \[ \lim_{n\to\infty}a_n \] exists.
Solution
Let \[ c\coloneqq\operatorname{Res}_{z=1}f(z). \]
1. The function\[ g(z)\coloneqq f(z)-\frac{c}{z-1} \]extends holomorphically to the whole disk \(\abs{z}<R\).
Because \(z=1\) is a simple pole of \(f\) with residue \(c\), its Laurent expansion near \(1\) has the form \[ f(z)=\frac{c}{z-1}+h(z), \] where \(h\) is holomorphic near \(1\). Thus \(g=h\) near the puncture, so the singularity of \(g\) at \(1\) is removable. Away from \(1\), both terms defining \(g\) are holomorphic on \(\abs{z}<R\).
2. Write\[ g(z)=\sum_{n=0}^{\infty}b_nz^n \]near \(0\). Then\[ b_n\longrightarrow0. \]
Choose any \(\rho\) with \[ 1<\rho<R. \] By step 1, \(g\) is holomorphic on a neighborhood of the closed disk \(\abs{z}\leq\rho\). Set \[ M_\rho\coloneqq\max_{\abs{z}=\rho}\abs{g(z)}. \] Cauchy’s coefficient formula gives \[ b_n = \frac{1}{2\pi i} \int_{\abs{z}=\rho} \frac{g(z)}{z^{n+1}}\,dz, \] and hence \[ \abs{b_n} \leq \frac{M_\rho}{\rho^n}. \] Since \(\rho>1\), the right-hand side tends to \(0\).
3. For every \(n\geq0\),\[ a_n=b_n-c. \]
For \(\abs{z}<1\), \[ \frac{c}{z-1} = -\frac{c}{1-z} = -c\sum_{n=0}^{\infty}z^n. \] Since \(f=g+c/(z-1)\), comparison with the Maclaurin series of \(f\) gives \(a_n=b_n-c\).
4. Therefore\[ \boxed{ \lim_{n\to\infty}a_n = -\operatorname{Res}_{z=1}f(z) }. \]
By step 3, \[ a_n=b_n-c, \] and step 2 gives \(b_n\to0\).
5. Q.E.D.
Step 4 proves, in particular, that the required limit exists.