Theorem (Hadamard's Formula). Let f:C→Cf : \mathbb{C} \rightarrow \mathbb{C} be a power series defined as f(z)=∑n≥1anznf(z) = \sum_{n \ge 1} a_n z^n. The radius of convergence of ff, denoted as R∈[0,∞]R \in [0, \infty], is given by

1/R=L≐lim sup⁡k≥1∣ak∣1/k. 1 / R = L \doteq \limsup_{k \ge 1} |a_k|^{1 / k}.

where we adopt the convention that 1/0=∞1 / 0 = \infty and 1/∞=01 / \infty = 0.

It is guaranteed that

  1. If ∣z∣<R|z| < R, then ff converges absolutely.
  2. If ∣z∣>R|z| > R, then ff diverges.

Remark: The region ∣z∣<R|z| < R is called the disc of convergence of ff.

Proof.
If ∣z∣<R|z| < R, then L⋅∣z∣<1L \cdot |z| < 1. Hence,

∃ϵ>0, r<1,  s.t.  (L+ϵ)⋅∣z∣<r. \exists \epsilon > 0, \, r < 1, \, \text{ s.t. } \, (L + \epsilon) \cdot |z| < r.

Since sup⁡k≥n∣ak∣1/k→L\sup_{k \ge n} | a_k |^{1 / k} \rightarrow L, for large enough nn, we have

L+ϵ≥sup⁡k≥n∣ak∣1/k≥∣an∣1/n. L + \epsilon \ge \sup_{k \ge n} | a_k |^{1 / k} \ge |a_n|^{1 / n}.

Therefore,

∣anzn∣=(∣an∣1/n⋅∣z∣)n≤((L+ϵ)⋅∣z∣)n≤rn. | a_n z^n | = \big( |a_n|^{1 / n} \cdot |z| \big)^n \le \big( (L + \epsilon) \cdot |z| \big)^n \le r^n.

It follows that ff converges absolutely.

If ∣z∣>R|z| > R, then L⋅∣z∣>1L \cdot |z| > 1. Hence,

∃ϵ>0, r>1,  s.t.  (L−ϵ)⋅∣z∣>r. \exists \epsilon > 0, \, r > 1, \, \text{ s.t. } \, (L - \epsilon) \cdot |z| > r.

Since sup⁡k≥n∣ak∣1/k\sup_{k \ge n} | a_k |^{1 / k} decreases to LL, for each n∈N+n \in \mathbb{N}^+,

∃k≥n,  s.t.  ∣ak∣1/k≥(L−ϵ). \exists k \ge n, \, \text{ s.t. } \, |a_k|^{1 / k} \ge (L - \epsilon).

For such kk,

∣akzk∣=(∣ak∣1/k⋅∣z∣)k≥((L−ϵ)⋅∣z∣)k≥rk. | a_k z^k | = \big( |a_k|^{1 / k} \cdot |z| \big)^k \ge \big( (L - \epsilon) \cdot |z| \big)^k \ge r^k.

Since the terms of ff are unbounded, the series is consequently divergent.

□\square

Theorem. Let f:C→Cf : \mathbb{C} \rightarrow \mathbb{C} be a power series defined as f(z)=∑n≥1anznf(z) = \sum_{n \ge 1} a_n z^n. Then it is holomorphic in its disc of convergence. Its derivative is given by

f′(z)=∑n≥1nanznn−1. f'(z) = \sum_{n \ge 1} n a_n z_n^{n - 1}.

Moreover, f′(z)f'(z) has the same disc of convergence.

Proof. Let bn=nanb_n = n a_n. Define

g(z)=∑n≥1bnznn−1=∑n≥1nanznn−1. g(z) = \sum_{n \ge 1} b_n z_n^{n - 1} = \sum_{n \ge 1} n a_n z_n^{n - 1}.

Observe that

lim sup⁡n∣bn∣1/n=lim sup⁡n∣nan∣1/n=lim⁡n∣n∣1/n⋅lim sup⁡n∣an∣1/n=lim sup⁡n∣an∣1/n. \begin{aligned} \limsup_n | b_n |^{ 1 / n} &= \limsup_n | n a_n |^{ 1 / n} \\ &= \lim_n | n |^{ 1 / n} \cdot \limsup_n | a_n |^{ 1 / n} \\ &= \limsup_n | a_n |^{ 1 / n}. \end{aligned}

It follows that g(z)g(z) has the same disc of convergence as f(z)f(z).

It is left to prove that g(z)g(z) is the derivative of f(z)f(z). For k∈N+k \in \mathbb{N}^+, write f(z)=pk(z)+rk(z)f(z) = p_k (z) + r_k(z), where

pk(z)=∑n∈[k]anzn,rk(z)=∑n>kanzn. p_k(z) = \sum_{n \in [k] } a_n z^n,\quad r_k(z) = \sum_{n > k} a_n z^n.

Denote RR the radius of convergence of f(z)f(z). and suppose 0≤∣z0∣<r<R0 \le |z_0| < r < R. Then for h∈Ch \in \mathbb{C}, such that ∣z0+h∣<r|z_0 + h| < r, we have

∣f(z0+h)−f(z0)h−g(z0)∣=∣pk(z0+h)−pk(z0)h+rk(z0+h)−rk(z0)h−g(z0)∣=∣rk(z0+h)−rk(z0)h+(pk(z0+h)−pk(z0)h−pk′(z0))+pk′(z0)−g(z0)∣≤∣pk(z0+h)−pk(z0)h−pk′(z0)∣+∣rk(z0+h)−rk(z0)h∣+∣pk′(z0)−g(z0)∣.. \begin{aligned} \left| \frac{ f(z_0 + h) - f(z_0) }{ h } - g(z_0) \right| &= \left| \frac{ p_k(z_0 + h) - p_k(z_0) }{ h } + \frac{ r_k(z_0 + h) - r_k(z_0) }{ h } - g(z_0) \right| \\ &= \left| \frac{ r_k(z_0 + h) - r_k(z_0) }{ h } + \left( \frac{ p_k(z_0 + h) - p_k(z_0) }{ h } - p_k'(z_0) \right) + p_k'(z_0) - g(z_0 ) \right| \\ &\le \left| \frac{ p_k(z_0 + h) - p_k(z_0) }{ h } - p_k'(z_0) \right| + \left| \frac{ r_k(z_0 + h) - r_k(z_0) }{ h } \right| + \left| p_k'(z_0) - g(z_0 ) \right| . \end{aligned}.

Claim. ∀ϵ>0\forall \epsilon > 0, ∃k∈N+\exists k \in \mathbb{N}^+, s.t.,

∣rk(z0+h)−rk(z0)h∣≤ϵ,∣pk′(z0)−g(z0)∣≤ϵ. \left| \frac{ r_k(z_0 + h) - r_k(z_0) }{ h } \right| \le \epsilon, \quad \left| p_k'(z_0) - g(z_0 ) \right| \le \epsilon.

Further, for such integer kk, ∃h>0\exists h > 0, s.t.,

∣pk(z0+h)−pk(z0)h−pk′(z0)∣≤ϵ. \left| \frac{ p_k(z_0 + h) - p_k(z_0) }{ h } - p_k'(z_0) \right| \le \epsilon.

Case One ∣rk(z0+h)−rk(z0)h∣≤ϵ\left| \frac{ r_k(z_0 + h) - r_k(z_0) }{ h } \right| \le \epsilon: since an−bn=(a−b)(an−1+an−2b2+…+bn−1),a^n - b^n = (a - b) \left( a^{n - 1} + a^{n - 2} b^2 + \ldots + b^{n - 1} \right), and ∣z0+h∣<r|z_0 + h| < r, ∣z0∣<r|z_0 | < r,

∣rk(z0+h)−rk(z0)h∣=∣∑n>kan[(z0+h)n−zn]h∣≤∑n>kannrn−1. \begin{aligned} \left| \frac{ r_k(z_0 + h) - r_k(z_0) }{ h } \right| &= \left| \frac{ \sum_{n > k} a_n \Big[ (z_0 + h)^n - z ^n \Big] }{ h } \right| &\le \sum_{n > k} a_n n r^{n - 1}. \end{aligned}

Since g(r)g(r) converges absolutely, for large enough kk,

∣rk(z0+h)−rk(z0)h∣≤∑n>kannrn−1≤ϵ. \left| \frac{ r_k(z_0 + h) - r_k(z_0) }{ h } \right| \le \sum_{n > k} a_n n r^{n - 1} \le \epsilon.

Case Two ∣pk′(z0)−g(z0)∣≤ϵ\left| p_k'(z_0) - g(z_0 ) \right| \le \epsilon: since g(z0)g(z_0) converges absolutely, for large enough kk,

∣pk′(z0)−g(z0)∣≤ϵ. \left| p_k'(z_0) - g(z_0 ) \right| \le \epsilon.

Case Three ∣pk(z0+h)−pk(z0)h−pk′(z0)∣≤ϵ\left| \frac{ p_k(z_0 + h) - p_k(z_0) }{ h } - p_k'(z_0) \right| \le \epsilon: following the definition of derivative, this is true for small enough hh.

□\square

Reference

[1] L. V. Ahlfors, Complex analysis: an introduction to the theory of analytic functions of one complex variable, 3d ed. New York: McGraw-Hill, 1979.