Let \(f\) be a power series centered at the origin. Prove that \(f\) has a power-series expansion around every point in its disc of convergence.
Solution
1. Write\[ f(z)=\sum_{n=0}^{\infty}a_nz^n \]with radius of convergence \(R\), and fix \(z_0\) with \(|z_0|<R\).
This is the hypothesis, with an arbitrary point chosen in the open disc of convergence.
2. If \(w\) satisfies \(|w|<R-|z_0|\), then\[ \sum_{n=0}^{\infty}|a_n|(|z_0|+|w|)^n<\infty. \]
The inequality \(|z_0|+|w|<R\) places the positive real number \(|z_0|+|w|\) strictly inside the radius of convergence. A power series converges absolutely at every point strictly inside its radius.
3. For such \(w\), the double series\[ \sum_{n=0}^{\infty}\sum_{k=0}^n |a_n|\binom nk|z_0|^{n-k}|w|^k \]converges.
By the binomial theorem, the inner sum equals \[ |a_n|(|z_0|+|w|)^n. \] Hence the double series is exactly the convergent series in 2.
4. Define, for \(k\ge0\),\[ c_k=\sum_{n=k}^{\infty}a_n\binom nk z_0^{\,n-k}. \]Then this series converges absolutely for every \(k\).
Choose any real \(\rho\) with \(|z_0|<\rho<R\). Since \(\sum |a_n|\rho^n<\infty\) and \[ \binom nk |z_0|^{n-k} =\rho^{-k}\binom nk\left(\frac{|z_0|}{\rho}\right)^{n-k}\rho^n, \] the factor \(\binom nk(|z_0|/\rho)^{n-k}\) is bounded in \(n\) for fixed \(k\): polynomial growth in \(n\) is dominated by the geometric factor \((|z_0|/\rho)^{n-k}\). Thus the defining series for \(c_k\) is dominated by a constant multiple of \(\sum |a_n|\rho^n\).
5. For every \(|w|<R-|z_0|\),\[ f(z_0+w)=\sum_{k=0}^{\infty}c_kw^k. \]
Using the binomial theorem, \[ \begin{aligned} f(z_0+w) &=\sum_{n=0}^{\infty}a_n(z_0+w)^n\\ &=\sum_{n=0}^{\infty}\sum_{k=0}^n a_n\binom nk z_0^{\,n-k}w^k. \end{aligned} \] By 3 this double series is absolutely convergent, so its terms may be rearranged. Summing first over \(n\ge k\) gives \[ f(z_0+w) =\sum_{k=0}^{\infty}\left(\sum_{n=k}^{\infty}a_n\binom nkz_0^{\,n-k}\right)w^k =\sum_{k=0}^{\infty}c_kw^k. \]
6. Therefore \(f\) has a power-series expansion centered at \(z_0\), valid at least on the disc \(|z-z_0|<R-|z_0|\).
5. Since \(z_0\) was arbitrary with \(|z_0|<R\), the conclusion holds around every point of the original disc of convergence.