Power series are functions defined by infinite algebraic data. Inside their radius of convergence they behave like polynomials: continuous, differentiable, and integrable term by term.
Radius of convergence
The convergence set of a power series is controlled by one number.
Theorem. For \(\sum_{n=0}^\infty a_n(x-c)^n\), there is \(R\in[0,\infty]\) such that the series converges absolutely for \(|x-c|\lt R\) and diverges for \(|x-c|>R\). Moreover \(1/R=\limsup |a_n|^{1/n}\).
Proof. Apply the root test to terms \(a_n(x-c)^n\). The limsup of the \(n\)-th roots is \(|x-c|\limsup |a_n|^{1/n}\). It is below \(1\) inside the radius and above \(1\) outside.
Remark (the boundary is separate data). The radius theorem says nothing at \(|x-c|=R\). For \(c=0\) and \(R=1\), the series \(\sum x^n\) diverges at both endpoints, \(\sum x^n/n\) converges at \(-1\) but diverges at \(1\), and \(\sum x^n/n^2\) converges at both. Endpoint behavior must therefore be checked directly.
Uniform convergence on compact subintervals
Inside the radius, convergence is locally uniform.
Proposition. If \(0\lt r\lt R\), then \(\sum a_n(x-c)^n\) converges uniformly and absolutely on \(|x-c|\le r\).
Proof. Choose \(\rho\) with \(r\lt \rho\lt R\). Convergence at \(c+\rho\) implies the terms \(a_n\rho^n\) are bounded, say \(|a_n|\rho^n\le M\). Therefore, on \(|x-c|\le r\), \[ |a_n(x-c)^n| \le M\left(\frac r\rho\right)^n. \] The geometric majorant is summable, so the Weierstrass test gives uniform absolute convergence.
Termwise differentiation
The derivative series has the same radius of convergence.
Theorem. Inside \(|x-c|\lt R\), the sum \(f(x)=\sum a_n(x-c)^n\) is differentiable and \[ f'(x)=\sum_{n=1}^\infty n a_n(x-c)^{n-1}. \]
Proof. The differentiated coefficients satisfy \[ \limsup_{n\to\infty}|na_n|^{1/n} =\left(\lim_{n\to\infty}n^{1/n}\right)\limsup_{n\to\infty}|a_n|^{1/n} =R^{-1}, \] so the differentiated series has radius \(R\). Fix \(r<R\). Its partial sums converge uniformly on \([c-r,c+r]\), while the original partial sums agree at \(c\). The uniform limit theorem for derivatives therefore yields differentiability of the sum and the asserted derivative. Repetition shows that \(f\in C^\infty(c-R,c+R)\) and \[ f^{(k)}(x)=\sum_{n=k}^{\infty}\frac{n!}{(n-k)!}a_n(x-c)^{n-k}. \]
Termwise integration
Power series can be integrated coefficient by coefficient.
Proposition. For \(|x-c|\lt R\), \[ \int_c^x\sum_{n=0}^\infty a_n(t-c)^n\,\mathrm{d}t =\sum_{n=0}^\infty \frac{a_n}{n+1}(x-c)^{n+1}. \]
Proof. On the interval between \(c\) and \(x\), choose a compact subinterval strictly inside the radius. Uniform convergence permits interchanging the Riemann or Lebesgue integral with the infinite sum.
Taylor coefficients and the identity principle
Coefficients are uniquely determined by the local function.
Corollary. If \(f(x)=\sum_{n=0}^{\infty}a_n(x-c)^n\) on \(|x-c|<R\), then \[ a_n=\frac{f^{(n)}(c)}{n!}. \] Consequently, two power series with the same sum on a neighborhood of their common center have identical coefficients.
Theorem. If a power series centered at \(c\) vanishes on a nonempty open interval inside its radius, then every coefficient is zero.
Proof. Let \(I=(c-R,c+R)\) and \[ Z=\{x\in I:f\text{ vanishes on a neighborhood of }x\}. \] The given open interval shows \(Z\ne\varnothing\), and \(Z\) is open by definition. To prove that \(Z\) is closed in \(I\), let \(x_j\in Z\) and \(x_j\to x\in I\). Every derivative \(f^{(m)}\) vanishes at every \(x_j\), hence continuity gives \(f^{(m)}(x)=0\) for all \(m\). A power series can be recentered at any interior point: for \(|h|<R-|x-c|\), absolute convergence and the binomial theorem give \[ f(x+h) =\sum_{m=0}^{\infty}\frac{f^{(m)}(x)}{m!}h^m. \] Thus \(f\) vanishes near \(x\), so \(x\in Z\). The interval \(I\) is connected; therefore \(Z=I\). In particular all \(f^{(n)}(c)\) vanish, and \(a_n=f^{(n)}(c)/n!=0\).
Example (smooth need not mean analytic). Define \[ g(x)= \begin{cases} e^{-1/x^2},&x\ne0,\\ 0,&x=0. \end{cases} \] Repeated differentiation shows \(g^{(k)}(x)=P_k(1/x)e^{-1/x^2}\) for \(x\ne0\), where \(P_k\) is a polynomial. Since \(t^m e^{-t^2}\to0\) as \(t\to\infty\), every derivative extends continuously to \(0\) with \(g^{(k)}(0)=0\). Its Taylor series at \(0\) is therefore identically zero although \(g(x)>0\) for \(x\ne0\). Power-series representability is strictly stronger than \(C^\infty\)-smoothness.