Power Series and Analyticity

We use the root test and a Weierstrass M-test argument; this is a direct comparison-with-geometric-series technique, entirely self-contained.

Part 1 (absolute convergence inside $R$). Let $z$ satisfy $|z - z_0| < R$. If $R = 0$ there is nothing to prove (no such $z$ exists). Assume $R > 0$, possibly $R = \infty$. Choose $\rho$ with $|z - z_0| < \rho < R$ (if $R = \infty$, choose any $\rho > |z-z_0|$). Since $\rho < R$, we have $1/\rho > 1/R = \limsup_n |a_n|^{1/n}$ when $R<\infty$; when $R=\infty$ the limsup is $0 < 1/\rho$ for any finite $\rho$. In either case, by definition of $\limsup$, there exists $N$ such that for all $n \geq N$,

Hence for $n \geq N$,

Since $|z-z_0|/\rho < 1$, this is a term of a convergent geometric series. By the comparison test, $\sum_{n \geq N} |a_n(z-z_0)^n|$ converges, and adding the finitely many terms $n < N$ (each a finite complex number) preserves convergence. Hence $\sum_n a_n(z-z_0)^n$ converges absolutely.

Part 2 (divergence outside $R$). Let $z$ satisfy $|z - z_0| > R$. If $R = \infty$ there is no such $z$. Assume $R < \infty$. Then $1/|z-z_0| < 1/R = \limsup_n |a_n|^{1/n}$. By definition of $\limsup$ as the supremum of subsequential limits, there is a subsequence $(n_k)$ along which $|a_{n_k}|^{1/n_k} \to L \geq 1/R$... more directly: since $\limsup_n |a_n|^{1/n} > 1/|z-z_0|$, there are infinitely many $n$ with

for infinitely many $n$. Thus the terms $a_n(z-z_0)^n$ do not converge to $0$, so the series $\sum a_n(z-z_0)^n$ diverges (a necessary condition for convergence of any series is that its terms tend to $0$).

Part 3 (uniform convergence on closed subdisks). Fix $0 < \rho < R$. Choose $\rho'$ with $\rho < \rho' < R$ (possible since $\rho < R$; if $R = \infty$ pick any $\rho' > \rho$). As in Part 1, there is $N$ such that $|a_n| < (\rho')^{-n}$ for all $n \geq N$. Then for every $z$ with $|z - z_0| \leq \rho$ and every $n \geq N$,

Since $\rho/\rho' < 1$, $\sum_n M_n$ is a convergent geometric series independent of $z$. By the Weierstrass M-test, $\sum_n a_n(z - z_0)^n$ converges uniformly on $\overline{D}(z_0,\rho)$.

Continuity. Each partial sum $S_N(z) = \sum_{n=0}^N a_n(z-z_0)^n$ is a polynomial, hence continuous on $\mathbb{C}$. A uniform limit of continuous functions on a set is continuous on that set (standard $\varepsilon/3$ argument: for $z, w$ in the disk, $|f(z)-f(w)| \le |f(z)-S_N(z)| + |S_N(z)-S_N(w)| + |S_N(w)-f(w)|$, and each piece is made small by uniform convergence and continuity of $S_N$). Since every point of $D(z_0,R)$ lies in some closed subdisk $\overline{D}(z_0,\rho)$ with $\rho<R$, the sum $f$ is continuous at every point of $D(z_0,R)$. $\blacksquare$

Same radius. Since $n^{1/n} \to 1$ as $n \to \infty$, we have $\limsup_n |na_n|^{1/n} = \limsup_n |a_n|^{1/n}$ (multiplying by a sequence tending to $1$ does not change a limsup: for any $\varepsilon>0$, eventually $n^{1/n} < 1+\varepsilon$ and $n^{1/n}>1-\varepsilon$, squeezing the limsup of the product between $(1\pm\varepsilon)$ times the original limsup, and letting $\varepsilon \to 0$). So $g$'s radius of convergence is also $R$ (index shift $n \to n+1$ does not affect the limsup either, by the same reasoning applied to the sequence $(n+1)a_{n+1}$).

Differentiability. Fix $z^* \in D(z_0,R)$ and choose $\rho$ with $|z^*-z_0| < \rho < R$. We show $f'(z^*) = g(z^*)$ directly from the difference quotient, using uniform convergence on $\overline D(z_0,\rho)$ established above (this is a term-by-term limit interchange argument). For $h \neq 0$ small enough that $z^*+h \in D(z_0,\rho)$, write

Let $w = z^* - z_0$, so $|w| < \rho$. Using the algebraic identity $u^n - v^n = (u-v)\sum_{k=0}^{n-1}u^kv^{n-1-k}$ with $u = w+h$, $v=w$, the bracket equals

Each term $(w+h)^k - w^k$ is again a telescoping sum bounded, for $|h|$ small enough that $|w+h| \le \rho$, by $k|h|\rho^{k-1}$ (apply the same factorization once more and bound each factor by $\rho^{k-1}$). Hence the bracket is bounded in absolute value by

Therefore

The series $\sum_n |a_n| n(n-1)\rho^{n-2}/2$ converges: it is (up to the same limsup argument as above, since $n(n-1)/2$ is polynomial in $n$ and $x^{1/n}\to1$ for any fixed polynomial factor $x=x(n)$) a power series in $\rho$ with the same radius of convergence $R$, evaluated at $\rho < R$, hence convergent by Part 1 of Radius of Convergence and the Cauchy–Hadamard Formula (the theorem above). Call this convergent sum $C < \infty$. Then

Hence $f$ is complex differentiable at $z^*$ with $f'(z^*) = g(z^*)$. Since $z^* \in D(z_0,R)$ was arbitrary, $f$ is holomorphic on $D(z_0,R)$ with derivative $g$. $\blacksquare$

($\Leftarrow$) Analytic implies holomorphic. This is immediate from the Lemma above: if $f$ is analytic at $z_0$, it agrees with a convergent power series on some disk $D(z_0,r)$, and that lemma shows such a series defines a holomorphic function on its disk of convergence. Since analyticity at every point of $U$ gives, at each point, a local power series representation with positive radius, $f$ is complex differentiable at every point of $U$, i.e. holomorphic.

($\Rightarrow$) Holomorphic implies analytic. This is the substantial direction; we use Cauchy's Integral Formula together with the geometric series and the uniform-convergence machinery above (a Cauchy kernel expansion argument). Fix $z_0 \in U$. Since $U$ is open, choose $R>0$ with $\overline{D}(z_0,R) \subseteq U$ (a closed disk of some positive radius fits inside $U$). Let $\gamma$ be the positively oriented circle $|\zeta - z_0| = R$, which lies in $U$ and is homotopic to a point in $U$, so Cauchy's Integral Formula applies to $f$ (holomorphic on an open set containing $\overline D(z_0,R)$) for any $z$ with $|z-z_0|<R$:

Fix such a $z$ and write $\zeta - z = (\zeta - z_0) - (z-z_0)$, so that

using the geometric series formula $\frac{1}{1-w} = \sum_{n\ge0}w^n$, valid since $\zeta \in \gamma$ gives $|\zeta - z_0| = R$ and $|z-z_0|<R$, so $|w| := \left|\frac{z-z_0}{\zeta-z_0}\right| = \frac{|z-z_0|}{R} < 1$. This series in $w$ converges uniformly for $\zeta \in \gamma$ (i.e. as $\zeta$ ranges over the circle, $|w|$ is the constant ratio $|z-z_0|/R < 1$, so this is literally a fixed convergent geometric series of numbers, uniform trivially in $\zeta$; more precisely, the partial sums converge uniformly in $\zeta$ by the Weierstrass M-test with $M_n = (|z-z_0|/R)^n$, a convergent geometric bound independent of $\zeta$). Therefore

and since $f$ is continuous on the compact set $\gamma$ (holomorphic functions are continuous, being complex differentiable), $f$ is bounded on $\gamma$, say $|f(\zeta)| \leq M$ for $\zeta \in \gamma$. The $n$-th term is then bounded by $M(|z-z_0|/R)^n/R$, uniformly in $\zeta \in \gamma$; the series of these bounds converges (geometric, ratio $<1$), so by the Weierstrass M-test the series $\sum_n \frac{f(\zeta)(z-z_0)^n}{(\zeta - z_0)^{n+1}}$ converges uniformly in $\zeta$ on $\gamma$. Uniform convergence permits interchange of the (contour) integral and the infinite sum — this is justified because a uniformly convergent series of continuous functions on a rectifiable curve of finite length may be integrated term by term (the partial sums converge uniformly to the full sum, and integration against a fixed finite-length curve is a continuous linear functional with respect to the sup norm, so it commutes with uniform limits). Hence

Define

These are well-defined complex numbers (independent of $z$), and we have shown

This exhibits $f$ as equal to a convergent power series on $D(z_0,R)$, so $f$ is analytic at $z_0$. Since $z_0 \in U$ was arbitrary, $f$ is analytic on all of $U$.

Identification of coefficients with Taylor coefficients. By the Lemma (termwise differentiation), applied repeatedly to $f(z) = \sum_n a_n(z-z_0)^n$ on $D(z_0,R)$, we get for every $k \geq 0$,

(each differentiation is justified by the Lemma, applied to the previous derivative series, which itself is a power series with the same radius $R$). Evaluating at $z = z_0$ kills every term except $n=k$, giving $f^{(k)}(z_0) = k! \, a_k$, i.e.

This shows the power series obtained from Cauchy's formula is exactly the Taylor series of $f$ at $z_0$, and it converges to $f$ on the disk $D(z_0,R)$ for every $R$ with $\overline{D}(z_0,R)\subseteq U$; taking the supremum of all such admissible $R$ recovers the statement for the largest disk $D(z_0,R)\subseteq U$ (any point of that open disk lies in some smaller closed disk contained in $U$, to which the above argument applies). $\blacksquare$