The Cauchy-Hadamard theorem

Theorem

The Cauchy-Hadamard theorem

T-143Home MU-104Threads change · space

Statement

Let \( (c_n)_{n \ge 0} \) be any sequence of complex numbers, let \( z_0 \in \mathbb{C} \), and consider the power series \( \sum_{n=0}^{\infty} c_n (z - z_0)^n \). Put \[ \alpha \;=\; \limsup_{n \to \infty} |c_n|^{1/n} \;\in\; [0, +\infty], \qquad R \;=\; \frac{1}{\alpha}, \] with the conventions \( 1/0 := +\infty \) and \( 1/(+\infty) := 0 \). Then: (i) for every \( z \) with \( |z - z_0| \lt R \) the series converges absolutely; (ii) for every \( z \) with \( |z - z_0| \gt R \) the terms \( c_n (z-z_0)^n \) are unbounded, so the series diverges; (iii) for every \( r \) with \( 0 \le r \lt R \) the convergence is uniform on the closed disc \( \{ z : |z - z_0| \le r \} \), so the sum function is continuous on \( |z - z_0| \lt R \). No claim whatever is made when \( |z - z_0| = R \). The number \( R \), the radius of convergence, is the unique element of \( [0, +\infty] \) with properties (i) and (ii); the same statement and proof hold verbatim for real coefficients and real \( z \), with discs replaced by intervals.

Why it matters

A power series is the basic constructive object of analysis: \( \exp \), \( \sin \), \( \log(1+z) \), the binomial series, Bessel functions, generating functions in combinatorics, perturbation expansions in physics and series solutions of differential equations are all defined by one. Before any of that can begin, one question must be answered — where does the series mean anything? Cauchy–Hadamard answers it completely and mechanically. The domain of convergence is always a disc, never a stranger shape, and its radius is read off the coefficients by a single formula that requires no hypothesis at all: the \( \limsup \) exists for every sequence, so the formula never fails to return an answer.

The formula also says what the radius means. Since \( R^{-1} = \limsup_n |c_n|^{1/n} \), the radius is precisely the reciprocal of the exponential growth rate of the coefficients: \( R \) is large exactly when \( |c_n| \) decays fast. That dictionary between coefficient size and domain size is the engine of the whole theory that follows — Cauchy's coefficient estimates, the identification of \( R \) with the distance from \( z_0 \) to the nearest singularity of the sum, truncation-error bounds in numerical analysis, and the modern asymptotic analysis of generating functions, where singularity location is the asymptotics. It is also the reason a real-variable phenomenon like the sudden failure of the series for \( 1/(1+x^2) \) at \( |x| = 1 \), where nothing visible goes wrong, has an explanation only in the complex plane.

Hypotheses

The exponent is \( n \), matching the index of \( c_n \).The formula reads growth of \( |c_n| \) against the \( n \)-th power of the distance, so the series must be presented as \( \sum c_n (z-z_0)^n \) with the same \( n \) upstairs and downstairs. For \( \sum_{n} a_n z^{2n} \) the correct move is to set \( u = z^2 \), get radius \( R_u = 1/\limsup |a_n|^{1/n} \) in \( u \), and then take a square root: the \( z \)-radius is \( \sqrt{R_u} \), not \( R_u \). Equivalently, expand to genuine coefficients \( c_{2n} = a_n \), \( c_{2n+1} = 0 \), and the formula gives \( \limsup_m |c_m|^{1/m} = \limsup_n |a_n|^{1/2n} = R_u^{-1/2} \) directly.

\( \limsup \), not \( \lim \).This is what makes the theorem unconditional. Take \( c_n = 2^n \) for \( n \) even and \( c_n = 1 \) for \( n \) odd. Then \( |c_n|^{1/n} \) alternates between \( 2 \) and \( 1 \) and has no limit, so any statement of the form “\( R = 1/\lim |c_n|^{1/n} \)” is simply undefined here — yet the series has a perfectly good radius, \( R = 1/2 \). Replacing \( \limsup \) by \( \liminf \) is worse than undefined: it is wrong, giving \( 1 \) instead of \( 1/2 \), and at \( z = 0.7 \) the even terms \( (2 \cdot 0.7)^{n} = 1.4^{n} \) blow up.

Strict inequalities \( |z - z_0| \lt R \) and \( |z - z_0| \gt R \).The circle \( |z - z_0| = R \) is genuinely outside the theorem's reach, and not for want of effort: \( \sum z^n \), \( \sum z^n/n \) and \( \sum z^n/n^2 \) all have \( R = 1 \) while diverging at every boundary point, converging at every boundary point except \( z = 1 \), and converging absolutely at every boundary point respectively. No formula in the coefficients' growth rate alone can separate these, because they have the same growth rate.

Uniform convergence only on \( |z - z_0| \le r \) with \( r \lt R \).Whether convergence is uniform on the whole open disc depends on the particular coefficients, and the theorem does not decide it. For \( \sum z^n \) with \( R = 1 \) it fails: the tail satisfies \( \sup_{|z| \lt 1} \bigl| \sum_{n \ge N} z^n \bigr| = +\infty \) for every \( N \), so the sup-norm of the tail does not go to zero. For \( \sum z^n/n^2 \), also with \( R = 1 \), it holds, by the M-test with \( M_n = 1/n^2 \) — but that is a fact about those coefficients, not a consequence of Cauchy–Hadamard, whose input cannot distinguish the two. What the theorem does give is normal convergence on compact subsets, which is exactly what (iii) says, since every compact subset of the open disc lies in some \( |z-z_0| \le r \).

The scalars form a complete field with the usual absolute value.Absolute convergence is turned into convergence by the Cauchy criterion, and that step needs completeness of \( \mathbb{C} \) (equivalently of \( \mathbb{R} \); see completeness of the reals). Over \( \mathbb{Q} \) with the ordinary absolute value the comparison in step 6 still bounds the partial sums, but they need not converge to a rational number, and the theorem's conclusion is false as stated: \( \sum_{n} z^n/n! \) at \( z = 1 \) has bounded, monotone rational partial sums with no rational limit.

Proof

Everything rests on two properties of the upper limit and on the convergence of a geometric series; no property of the coefficients is used, and no continuity or differentiability is available at this stage. Throughout, \( \alpha = \limsup_n |c_n|^{1/n} \) and \( R = 1/\alpha \) with the stated conventions.

Substitute \( w = z - z_0 \). It suffices to prove the theorem for \[ \sum_{n=0}^{\infty} c_n w^n, \qquad w \in \mathbb{C}, \] since \( z \mapsto z - z_0 \) is a bijection of \( \mathbb{C} \) carrying \( \{ |z - z_0| \lt R \} \) onto \( \{ |w| \lt R \} \) and preserving every modulus \( |z - z_0| = |w| \).

A translation changes neither the terms of the series nor the coefficients, so the radius is a property of \( (c_n) \) alone; the centre only says where the disc sits. A

The quantity \( \alpha \) is well defined in \( [0, +\infty] \): writing \( x_n = |c_n|^{1/n} \ge 0 \) for \( n \ge 1 \) and \[ s_N \;=\; \sup_{n \ge N} x_n \;\in\; [0, +\infty], \] the sequence \( (s_N)_{N \ge 1} \) is nonincreasing, and \( \alpha = \lim_{N \to \infty} s_N = \inf_{N \ge 1} s_N \).

Each \( s_{N+1} \) is a supremum over a subset of the set defining \( s_N \), hence \( s_{N+1} \le s_N \); a nonincreasing sequence in \( [0,+\infty] \) converges to its infimum (see the monotone convergence theorem, extended to \( +\infty \) by declaring \( s_N \to +\infty \) when every \( s_N = +\infty \)). This is the whole reason the theorem needs no hypothesis: \( \alpha \) always exists. A

Two comparison properties of \( \alpha \), the only ones used below. For \( t \in (0, +\infty) \): \[ \begin{cases} t \gt \alpha & \Longrightarrow\ \exists N \ \forall n \ge N: \ |c_n|^{1/n} \lt t, \\ t \lt \alpha & \Longrightarrow\ |c_n|^{1/n} \gt t \ \text{ for infinitely many } n. \end{cases} \]

First line: if \( t \gt \alpha = \inf_N s_N \) then \( t \) is not a lower bound, so \( s_N \lt t \) for some \( N \), and \( s_N \) dominates every \( x_n \) with \( n \ge N \). Second line: if only finitely many \( n \) had \( x_n \gt t \) then \( s_N \le t \) for \( N \) past the last one, forcing \( \alpha \le t \), a contradiction. Both lines are valid when \( \alpha = 0 \) (second vacuous) and when \( \alpha = +\infty \) (first vacuous). B

Fix \( w \) with \( 0 \lt |w| \lt R \) and choose an intermediate radius \[ \rho \;=\; \begin{cases} \tfrac{1}{2}\bigl( |w| + R \bigr), & R \lt +\infty, \\ |w| + 1, & R = +\infty, \end{cases} \qquad \text{so that } \ |w| \lt \rho \lt R \ \text{ and } \ \frac{1}{\rho} \gt \frac{1}{R} = \alpha . \] Set \( \theta = |w|/\rho \), so \( 0 \lt \theta \lt 1 \).

The midpoint of \( (|w|, R) \) lies strictly inside it because \( |w| \lt R \); reciprocals reverse strict inequalities among positive numbers, and \( \rho \lt R \) gives \( 1/\rho \gt 1/R \). The case \( w = 0 \) is trivial (all terms after the first vanish), and if \( R = 0 \) there is no such \( w \) and (i) is vacuous. This choice of a margin \( \rho \) strictly between \( |w| \) and \( R \) is the single idea of the proof. B

Apply step 3 with \( t = 1/\rho \gt \alpha \): there is \( N \) with \[ |c_n|^{1/n} \lt \frac{1}{\rho} \quad \text{for all } n \ge N, \qquad \text{i.e.} \qquad |c_n| \lt \rho^{-n} \quad (n \ge N). \]

Raising both sides of \( |c_n|^{1/n} \lt \rho^{-1} \) to the \( n \)-th power is legitimate because \( x \mapsto x^n \) is strictly increasing on \( [0,\infty) \). A

Hence for all \( n \ge N \) \[ |c_n w^n| \;=\; |c_n| \, |w|^n \;\lt\; \frac{|w|^n}{\rho^n} \;=\; \theta^n, \qquad \text{so} \qquad \sum_{n = N}^{M} |c_n w^n| \;\lt\; \sum_{n=N}^{M} \theta^n \;\lt\; \frac{\theta^{N}}{1 - \theta} \] for every \( M \ge N \).

Multiplicativity of the modulus, then the bound of step 5, then the finite geometric sum \( \sum_{n=N}^{M} \theta^n = \theta^N \frac{1 - \theta^{M-N+1}}{1-\theta} \le \theta^N/(1-\theta) \), valid since \( 0 \lt \theta \lt 1 \). The whole content of the theorem is here: inside the margin, the terms are trapped under a fixed convergent geometric series. B

The partial sums of \( \sum_n |c_n w^n| \) are nondecreasing and bounded above by \( \sum_{n=0}^{N-1} |c_n w^n| + \theta^N/(1-\theta) \), hence convergent; so \( \sum_n c_n w^n \) converges absolutely, and therefore converges. This proves (i).

Nondecreasing because the summands are \( \ge 0 \); bounded by splitting off the finitely many terms below \( N \) and applying step 6. Convergence of a bounded monotone real sequence is the monotone convergence theorem, and the passage from absolute convergence to convergence is the Cauchy criterion together with \( \bigl| \sum_{n=p}^{q} c_n w^n \bigr| \le \sum_{n=p}^{q} |c_n w^n| \) — both statements about \( \mathbb{R} \) and \( \mathbb{C} \) that rest on completeness. A

Now fix \( r \) with \( 0 \lt r \lt R \) (for \( r = 0 \) the disc is the single point \( w = 0 \) and (iii) is trivial) and run steps 4–6 with \( |w| \) replaced by \( r \): with \( \rho \in (r, R) \) and \( \theta = r/\rho \in (0,1) \) there is \( N \) such that \[ \sup_{|w| \le r} |c_n w^n| \;=\; |c_n| \, r^n \;\lt\; \theta^n \quad (n \ge N), \qquad \sum_{n \ge N} \theta^n \lt \infty, \] so \( \sum_n c_n w^n \) converges uniformly — indeed normally — on \( \{ |w| \le r \} \). This proves (iii), and the sum is continuous there as a uniform limit of polynomials.

The supremum of \( |c_n w^n| \) over the closed disc is attained on its boundary because \( |w| \mapsto |c_n| |w|^n \) is nondecreasing; the summable dominating sequence \( M_n = \theta^n \) is exactly the hypothesis of the Weierstrass M-test, whose conclusion is uniform convergence, and a uniform limit of continuous functions is continuous. Every compact \( K \subset \{|w| \lt R\} \) satisfies \( K \subseteq \{|w| \le r\} \) with \( r = \max_{w \in K} |w| \lt R \), so this gives local uniform convergence on the whole open disc. B

Fix \( w \) with \( |w| \gt R \) and choose \( \rho \) with \( R \lt \rho \lt |w| \); then \( 1/\rho \lt \alpha \). By step 3 there are infinitely many \( n \) with \( |c_n| \gt \rho^{-n} \), and for those \( n \) \[ |c_n w^n| \;\gt\; \left( \frac{|w|}{\rho} \right)^{\! n} \;=\; \Theta^n, \qquad \Theta \;=\; \frac{|w|}{\rho} \;\gt\; 1 . \] Since \( \Theta^n \to +\infty \), the terms are unbounded; in particular \( c_n w^n \not\to 0 \) and the series diverges. This proves (ii).

Such a \( \rho \) exists: take \( \tfrac12 (R + |w|) \) if \( R \lt \infty \), and if \( R = 0 \) any \( \rho \in (0, |w|) \). The inequality \( 1/\rho \lt 1/R = \alpha \) again reverses under reciprocals. The last implication is the vanishing criterion: if \( \sum a_n \) converges then \( a_n = S_n - S_{n-1} \to S - S = 0 \), so unbounded terms forbid convergence. Note the conclusion is stronger than divergence — the terms blow up along a subsequence. B

\( R \) is the unique element of \( [0,+\infty] \) satisfying (i) and (ii), and moreover \[ R \;=\; \sup \bigl\{\, |w| \;:\; (c_n w^n)_{n \ge 0} \ \text{is bounded} \,\bigr\} \;=\; \sup \bigl\{\, |w| \;:\; \textstyle\sum_n c_n w^n \ \text{converges} \,\bigr\} . \]

Uniqueness: if \( R' \lt R \) satisfied (ii) it would force divergence at some \( w \) with \( R' \lt |w| \lt R \), contradicting (i) for \( R \); symmetrically for \( R' \gt R \). For the two suprema, call them \( B \) and \( S \). Step 9 gives \( B \le R \) and \( S \le R \), since terms are unbounded beyond \( R \); step 7 gives \( S \ge R \), since the series converges (so its terms are bounded, giving \( B \ge R \) too) at every \( |w| \lt R \). Hence \( B = S = R \). The middle characterisation is Abel's lemma in disguise: mere boundedness of the terms at one point \( w_0 \) forces absolute convergence on \( |w| \lt |w_0| \), because \( |c_n w^n| \le \bigl(\sup_k |c_k w_0^k| \bigr) \, (|w|/|w_0|)^n \). C

Result

\[ R \;=\; \frac{1}{\displaystyle \limsup_{n \to \infty} |c_n|^{1/n}}, \qquad \sum_{n=0}^{\infty} c_n (z-z_0)^n \ \ \begin{cases} \text{converges absolutely}, & |z - z_0| \lt R, \\ \text{diverges (unbounded terms)}, & |z - z_0| \gt R. \end{cases} \]

Reading. A power series converges exactly on a disc, and the radius of that disc is the reciprocal of the exponential growth rate of its coefficients. Coefficients that decay like \( \sigma^{n} \) buy a radius \( 1/\sigma \); coefficients that grow like \( n^n \) buy nothing at all; coefficients that decay like \( 1/n! \) buy the whole plane. Inside the disc the convergence is absolute and locally uniform, so the sum is a continuous function there; outside, the individual terms explode.

Scope. Valid for arbitrary complex (or real) coefficients with no assumption whatever — zero coefficients, gaps, oscillating growth, all permitted — because \( \limsup \) always exists in \( [0,+\infty] \). It says nothing about the circle \( |z-z_0| = R \), where all three of convergence everywhere, divergence everywhere, and mixed behaviour genuinely occur, and it does not assert uniform convergence on the full open disc, only on each closed sub-disc of strictly smaller radius.

Corollaries & converses

The domain of convergence is a disc. The set where a power series converges is squeezed between the open disc of radius \( R \) and its closure; it is never an annulus, a half-plane, or a scattered set. Any question about “where the series lives” reduces to one number plus a separate boundary analysis.
Ratio corollary (d'Alembert). If \( c_n \neq 0 \) eventually and \( |c_{n+1}/c_n| \to \ell \in [0,+\infty] \), then \( \limsup_n |c_n|^{1/n} = \ell \) and hence \( R = 1/\ell \). This is the version used in practice; it is a special case, and it is strictly weaker, since the limit of ratios need not exist (see the ratio test and Example 1).
Abel's lemma. If \( (c_n w_0^n) \) is bounded for a single \( w_0 \neq 0 \), then \( R \ge |w_0| \) and the series converges absolutely on \( |z - z_0| \lt |w_0| \). Conversely, unboundedness of \( (c_n w_0^n) \) forces \( R \le |w_0| \). Boundedness at one point is as good as convergence at that point.
Termwise differentiation and integration preserve \( R \). Because \( n^{1/n} \to 1 \), the series \( \sum n c_n (z - z_0)^{n-1} \) and \( \sum \frac{c_n}{n+1} (z - z_0)^{n+1} \) have the same radius as \( \sum c_n (z-z_0)^n \). Boundary behaviour is not preserved (Problem 4).
Gaps are free. Zero coefficients cause no trouble: a lacunary series such as \( \sum_k z^{2^k} \) has \( |c_n|^{1/n} \in \{0, 1\} \) and so \( R = 1 \). No ratio criterion can be formulated for it, because \( c_{n+1}/c_n \) is undefined infinitely often.
Every radius is realised. \( c_n = n^n \) gives \( R = 0 \); \( c_n = a^{-n} \) gives \( R = a \) for each \( a \in (0,\infty) \); \( c_n = 1/n! \) or \( c_n = n^{-n} \) gives \( R = +\infty \). So \( [0,+\infty] \) is exactly the set of possible radii.
Coefficient estimates. Rewriting the formula: for every \( \varepsilon \gt 0 \), \( |c_n| \le (R^{-1} + \varepsilon)^n \) for all large \( n \), and \( |c_n| \ge (R^{-1} - \varepsilon)^n \) for infinitely many \( n \). The upper half is the elementary shadow of Cauchy's estimates \( |c_n| \le M(r)/r^n \) from complex analysis; the lower half is what makes \( R \) sharp rather than merely a lower bound.
Sums and products. \( R_{f+g} \ge \min(R_f, R_g) \), with equality whenever \( R_f \neq R_g \); the Cauchy product also has radius at least \( \min(R_f, R_g) \). Strict inequality is possible when the radii coincide: \( f = \sum_n z^n \) and \( g = \sum_n (2^{-n} - 1) z^n \) both have radius \( 1 \), yet \( f + g = \sum_n 2^{-n} z^n \) has radius \( 2 \) — the leading coefficient growth cancels.
Converse fails in the boundary direction. \( R \) determines the open disc of convergence completely and the circle not at all; two series with identical \( R \) may converge at all, some, or none of the boundary points (Problem 5).

Fails without

\( \limsup \) replaced by \( \lim \): for \( c_n = 2^n \) (\( n \) even), \( c_n = 1 \) (\( n \) odd), \( |c_n|^{1/n} \) alternates \( 2, 1, 2, 1, \dots \) and has no limit, so the “limit form” of the formula returns nothing. The theorem as stated still applies and gives \( R = 1/2 \), confirmed directly: at \( |z| = 0.4 \) the terms are at most \( 0.8^n \), and at \( |z| = 0.7 \) the even terms are \( 1.4^{n} \to \infty \).
\( \limsup \) replaced by \( \liminf \): the same sequence has \( \liminf_n |c_n|^{1/n} = 1 \), predicting \( R = 1 \). That prediction is false: the series diverges at \( z = 0.7 \), a point strictly inside the predicted disc. \( \liminf \) yields only the statement that the terms are small along a subsequence, which is worthless for convergence.
Ratio in place of root: for \( c_n = 1 \) when \( n \) is a perfect square and \( c_n = 0 \) otherwise, the ratio \( c_{n+1}/c_n \) is undefined for infinitely many \( n \), so d'Alembert's corollary cannot even be stated; the root formula gives \( \limsup |c_n|^{1/n} = 1 \), hence \( R = 1 \), with no difficulty. Whenever the coefficients have gaps or oscillating scale, the ratio version fails and the \( \limsup \) version does not.
Strict inequality dropped at \( |z - z_0| = R \): \( \sum_{n \ge 0} z^n \), \( \sum_{n \ge 1} z^n/n \) and \( \sum_{n \ge 1} z^n/n^2 \) all have \( R = 1 \), but on \( |z| = 1 \) the first diverges everywhere, the second converges everywhere except at \( z = 1 \), and the third converges absolutely everywhere. Any conclusion drawn at radius exactly \( R \) is refuted by one of the three.
Uniform convergence claimed on the open disc: for \( \sum z^n \) the tail from \( N \) is \( z^N/(1-z) \), whose supremum over \( |z| \lt 1 \) is \( +\infty \) for every \( N \), so convergence is not uniform on \( |z| \lt 1 \) even though it is uniform on every \( |z| \le r \lt 1 \). The hypothesis \( r \lt R \) in (iii) cannot be relaxed to \( r = R \).
Completeness dropped: over \( \mathbb{Q} \) the estimate of step 6 still bounds the partial sums, but step 7's monotone-convergence and Cauchy-criterion arguments have no limit to produce. \( \sum_{n} 1/n! \) has rational partial sums, bounded above by \( 3 \), that converge to no rational number.

Common errors

“\( R = \lim_n |c_n / c_{n+1}| \) always.” Only when that limit exists. It is a corollary with an extra hypothesis, not the theorem; Example 1 and any lacunary series break it.
“Take the \( n \)-th root of \( c_n \).” The formula is \( |c_n|^{1/n} \): modulus first, root second. For complex \( c_n \) the expression \( c_n^{1/n} \) is multivalued and does not have a limit superior at all; taking moduli first makes every quantity a nonnegative real.
“The series converges on \( |z - z_0| \le R \).” The circle is undecided by this theorem and must be checked point by point with other tools — the alternating series test, Dirichlet's test, or absolute convergence.
“Convergence is uniform where the series converges.” What the theorem supplies is uniformity on each closed sub-disc of radius \( r \lt R \); on the full open disc it may hold (as for \( \sum z^n/n^2 \)) or fail (as for \( \sum z^n \)), and nothing in \( R \) alone tells you which. Statements that need uniformity — termwise integration, interchanging limits, continuity of the sum — must be applied on such a sub-disc first and then let \( r \uparrow R \).
Re-indexing errors. For \( \sum a_n z^{2n} \) the radius is \( \bigl( \limsup |a_n|^{1/n} \bigr)^{-1/2} \), not \( \bigl( \limsup |a_n|^{1/n} \bigr)^{-1} \); for \( \sum c_n (z - z_0)^n \) the disc is centred at \( z_0 \), so \( |z| \lt R \) is the wrong region unless \( z_0 = 0 \).
“\( \sum c_n z^n \) diverges outside \( R \), so the function is undefined there.” The series diverges; the function it represents inside the disc often continues analytically far beyond it. \( \sum z^n \) diverges at \( z = 2 \), while \( 1/(1-z) \) is perfectly well behaved there. Radius of convergence is a fact about the series at a centre, not about the function.
Concluding “convergence” only. Inside the disc the convergence is absolute, hence unconditional: rearrangements, Cauchy products and termwise manipulations are all licensed. That extra strength is free and is used constantly.

Discussion

The formula is usually credited jointly: it appears in Cauchy's Cours d'analyse of 1821, alongside his root test for series, and was then largely overlooked until Hadamard rediscovered it and made it the organising principle of his work on Taylor series and their singularities around 1888–1892. The double attribution is fair, because the two authors used it differently. For Cauchy it was a convergence criterion; for Hadamard it was a measuring instrument, a way of extracting analytic information about a function — where its nearest singularity sits, how the singularities are distributed on the circle — from the arithmetic of its coefficients. Everything in modern analytic combinatorics, where the asymptotics of a counting sequence are read off the singularities of its generating function, is a descendant of that shift in viewpoint.

The proof shows why \( \limsup \) is not merely a technical convenience but the correct object. Convergence of \( \sum c_n w^n \) is decided by a competition between the growth of \( |c_n| \) and the decay of \( |w|^n \), and only the worst coefficients matter: one subsequence of large \( |c_n| \) is enough to destroy convergence, while sporadically small coefficients help nobody. “Eventually below \( t \)” and “infinitely often above \( t \)” are exactly the two halves of the \( \limsup \), and they are exactly the two halves of the theorem — step 3 is not a lemma the proof happens to need, it is the theorem in embryo. Anything less than \( \limsup \) loses one of the halves: \( \lim \) may not exist, and \( \liminf \) controls the wrong subsequence.

The most striking consequence is a real-variable one that only the complex statement explains. The function \( f(x) = 1/(1+x^2) \) is infinitely differentiable on all of \( \mathbb{R} \), bounded by \( 1 \), with no visible defect anywhere; yet its Maclaurin series \( \sum (-1)^n x^{2n} \) has radius \( 1 \) and is useless for \( |x| \gt 1 \). Cauchy–Hadamard computes the radius but does not, by itself, explain it. The explanation is that the coefficients remember the poles at \( z = \pm i \), which sit at distance \( 1 \) from the origin in a plane the real graph cannot see. Once complex analysis is available the identification is exact: the radius of the Taylor series of a holomorphic function at \( z_0 \) equals the distance from \( z_0 \) to the nearest point where the function fails to be holomorphic. Cauchy–Hadamard is the elementary, coefficient-side half of that statement, and it is the half that requires no analyticity at all.

Two refinements mark where the elementary theory stops. First, the boundary is not merely undecided by the formula; it is undecidable in principle from growth rates, since \( \sum z^n/n \) and \( \sum z^n \) have identical \( \limsup |c_n|^{1/n} = 1 \) and opposite boundary behaviour. What can be said needs finer tools: Abel's theorem (if the series converges at a boundary point \( \zeta \), the sum tends to that value radially as \( z \to \zeta \)), Tauberian converses of it, and Fatou-type results relating boundary convergence to the analytic continuation. Second, whenever \( 0 \lt R \lt \infty \) the circle \( |z - z_0| = R \) must carry at least one genuine singularity of the sum, for otherwise the function would extend holomorphically to a strictly larger disc and its Taylor series would have a larger radius; when the coefficients are nonnegative, Pringsheim's theorem locates such a singularity at the real point \( z_0 + R \). In extreme cases every boundary point is one. The Ostrowski–Hadamard gap theorem makes this precise: if \( f(z) = \sum_k a_k z^{n_k} \) has radius \( 1 \) and the exponents are lacunary in the sense that \( n_{k+1}/n_k \ge \lambda \gt 1 \), the unit circle is a natural boundary and \( f \) admits no analytic continuation past any point of it. Problem 3 constructs such a series, whose sum is continuous on the entire closed unit disc and yet cannot be extended past a single boundary point — a good corrective to the intuition that convergence at the boundary means the function “keeps going”.

Common misconceptions. The radius of convergence is not a property of the function but of the series at a chosen centre: \( 1/(1-z) \) has radius \( 1 \) about \( 0 \) and radius \( 3 \) about \( z_0 = -2 \). A large radius is not a statement that the series is useful at large \( |z| \) — convergence can be intolerably slow long before the boundary, which is why truncation error, governed by the ratio \( |z - z_0| / R \), matters more in practice than \( R \) itself. And \( R = 0 \) does not make a series meaningless: divergent series such as \( \sum n!\, z^n \) carry real asymptotic information, and the whole theory of asymptotic (Borel-summable) expansions exists to extract it.

Worked examples

Example 1 (a series with no ratio limit). Find the radius and disc of convergence of \[ \sum_{n=0}^{\infty} c_n (z - 2)^n, \qquad c_n = \begin{cases} 2^{-n}, & n \text{ even}, \\ 3^{-n}, & n \text{ odd}, \end{cases} \] show that the ratio corollary is inapplicable, and confirm the answer against a closed form for the sum.

The ratio corollary fails. With \( w = z - 2 \) and \( n \) even, \[ \left| \frac{c_{n+1}}{c_n} \right| = \frac{3^{-(n+1)}}{2^{-n}} = \frac{1}{3} \left( \frac{2}{3} \right)^{\! n} \longrightarrow 0, \] while for \( n \) odd \[ \left| \frac{c_{n+1}}{c_n} \right| = \frac{2^{-(n+1)}}{3^{-n}} = \frac{1}{2} \left( \frac{3}{2} \right)^{\! n} \longrightarrow +\infty . \]

Direct computation of the two subsequences of ratios. Since one subsequence tends to \( 0 \) and the other to \( +\infty \), \( \lim_n |c_{n+1}/c_n| \) does not exist and d'Alembert's corollary says nothing at all. A

The root quantities are constant along each parity class: \[ |c_n|^{1/n} = \begin{cases} (2^{-n})^{1/n} = \tfrac{1}{2}, & n \text{ even}, \\[2pt] (3^{-n})^{1/n} = \tfrac{1}{3}, & n \text{ odd}. \end{cases} \]

\( (a^{-n})^{1/n} = a^{-1} \) for \( a \gt 0 \). Taking the \( n \)-th root is exactly what flattens the two competing geometric scales into two constants. A

Therefore \[ \alpha = \limsup_{n \to \infty} |c_n|^{1/n} = \max\left\{ \tfrac{1}{2}, \tfrac{1}{3} \right\} = \tfrac{1}{2}, \qquad R = \frac{1}{\alpha} = 2 . \]

For a sequence taking only the two values \( \tfrac12 \) and \( \tfrac13 \), each infinitely often, every tail supremum \( s_N \) equals \( \tfrac12 \), so the infimum of the tail suprema is \( \tfrac12 \). Only the larger value survives — the worst coefficients set the radius. B

Numerical check at two points. At \( z = 3.5 \) (so \( |w| = 1.5 \lt 2 \)) the terms are at most \( (1.5/2)^n = 0.75^n \) and the series converges; the partial sums \( S_{10} = 2.8793 \), \( S_{20} = 2.9483 \), \( S_{25} = 2.9511 \) climb towards \( 62/21 = 2.95238\ldots \). At \( z = 4.5 \) (so \( |w| = 2.5 \gt 2 \)) the even terms are \( 2^{-n} 2.5^{\,n} = 1.25^{\,n} \), which at \( n = 40 \) already exceeds \( 7.5 \times 10^{3} \).

The first computation is step 6 of the proof with \( \rho = 2 \) and \( \theta = 0.75 \); the second is step 9 with \( \rho = 2 \) and \( \Theta = 1.25 \gt 1 \), showing unbounded terms and hence divergence. A

Closed form, for \( |w| \lt 2 \): splitting into even and odd indices, \[ f(w) = \sum_{m \ge 0} \left( \frac{w}{2} \right)^{2m} \! + \sum_{m \ge 0} \left( \frac{w}{3} \right)^{2m+1} = \frac{1}{1 - w^2/4} + \frac{w/3}{1 - w^2/9} = \frac{4}{4 - w^2} + \frac{3w}{9 - w^2} . \] Its singularities are at \( w = \pm 2 \) and \( w = \pm 3 \); the nearest are at distance \( 2 \).

Rearranging into two subseries is licensed by absolute convergence (step 7); each is geometric with ratio \( w^2/4 \) and \( w^2/9 \) respectively, both of modulus \( \lt 1 \) when \( |w| \lt 2 \). At \( w = 1.5 \), \( f = \tfrac{16}{7} + \tfrac{2}{3} = \tfrac{62}{21} \), matching the partial sums of step 4. The distance \( 2 \) from the centre to the nearest singularity agrees with \( R \), as the complex-analytic identification predicts. B

\[ R = 2, \qquad \text{convergence exactly on } \ |z - 2| \lt 2 \ \ \bigl( \text{real interval } (0,4) \bigr), \qquad \text{divergence for } |z-2| \gt 2. \]

Reading. Two interleaved geometric scales, \( 2^{-n} \) and \( 3^{-n} \), compete; the \( \limsup \) picks the slower decay, \( 2^{-n} \), and the radius is set by it alone. The odd coefficients, being smaller, are irrelevant to the radius even though they are half of the series.

Scope. Boundary points \( z = 0 \) and \( z = 4 \) are not decided by the theorem; direct inspection shows the even terms equal \( 1 \) there, so the terms do not tend to \( 0 \) and the series diverges at both. The example is the standard demonstration that Cauchy–Hadamard strictly outperforms the ratio test.

Example 2 (the relativistic \( \gamma \)-factor, with units). The Lorentz factor of special relativity is \( \gamma = (1 - \beta^2)^{-1/2} \) with \( \beta = v/c \), where \( c = 299\,792\,458\ \mathrm{m\,s^{-1}} \) exactly (the SI definition of the metre). Its binomial expansion is \[ \gamma \;=\; \sum_{n=0}^{\infty} \binom{2n}{n} \frac{u^n}{4^n}, \qquad u = \beta^2 . \] Find the radius of convergence in \( u \), translate it into a speed, and evaluate the kinetic energy \( K = (\gamma - 1) m c^2 \) of a proton at \( v = 1.00 \times 10^{8}\ \mathrm{m\,s^{-1}} \) from the series, using the standard proton mass \( m_{p} = 1.6726 \times 10^{-27}\ \mathrm{kg} \).

Bound the coefficients \( a_n = \binom{2n}{n} 4^{-n} \) between two elementary expressions: \[ \frac{1}{2n+1} \;\le\; a_n \;\le\; 1 \qquad (n \ge 0). \]

Upper bound: \( \binom{2n}{n} \le \sum_{k=0}^{2n} \binom{2n}{k} = 4^n \). Lower bound: \( \binom{2n}{n} \) is the largest of the \( 2n+1 \) binomial coefficients \( \binom{2n}{k} \), so \( (2n+1)\binom{2n}{n} \ge 4^n \). No Stirling formula is needed; two counting inequalities suffice. B

Take \( n \)-th roots and squeeze: \[ \left( \frac{1}{2n+1} \right)^{\! 1/n} \le a_n^{1/n} \le 1, \qquad \left( \frac{1}{2n+1} \right)^{\! 1/n} = \exp\!\left( - \frac{\ln (2n+1)}{n} \right) \longrightarrow e^{0} = 1 . \] Hence \( \lim_n a_n^{1/n} = 1 \), so \( \alpha = \limsup_n a_n^{1/n} = 1 \) and the radius in \( u \) is \( R_u = 1 \).

The map \( x \mapsto x^{1/n} \) is increasing on \( [0,\infty) \); \( \ln(2n+1)/n \to 0 \) because logarithms grow more slowly than any positive power; the squeeze theorem then forces the limit, and when a limit exists it equals the \( \limsup \). A

Translate to speed. Convergence needs \( |u| = \beta^2 \lt 1 \), i.e. \[ \frac{|v|}{c} \lt 1 \qquad \Longleftrightarrow \qquad |v| \lt c = 2.99792458 \times 10^{8}\ \mathrm{m\,s^{-1}} . \]

\( \beta^2 \lt 1 \) with \( \beta \) real is equivalent to \( |\beta| \lt 1 \); multiplying by the positive constant \( c \) preserves the inequality and restores the units \( \mathrm{m\,s^{-1}} \). The expansion therefore has exactly the physical domain it should: every sub-luminal speed, and nothing beyond. A

Insert the data. \[ \beta = \frac{1.00 \times 10^{8}}{2.99792458 \times 10^{8}} = 0.333564, \qquad u = \beta^2 = 0.111265 \lt 1 . \] Partial sums of the series: \[ 1, \quad 1.055633, \quad 1.060275, \quad 1.060705, \quad 1.060747, \quad 1.060752, \] against the exact value \( \gamma = (1 - u)^{-1/2} = 1.0607520 \).

Each new term is roughly \( u \approx 0.111 \) times the previous one — the geometric decay of step 6 of the proof with \( \theta \approx |u|/\rho \) for any \( \rho \) between \( 0.111 \) and \( 1 \). Five terms already give six-figure agreement; the truncation error after \( N \) terms is \( \mathcal{O}(u^{N}) \), which is why the expansion is a practical tool at this speed and useless near \( \beta = 1 \). B

Assemble the energy, symbols first: \[ K = (\gamma - 1)\, m_p c^2, \qquad m_p c^2 = \bigl( 1.6726 \times 10^{-27}\ \mathrm{kg} \bigr) \bigl( 2.99792458 \times 10^{8}\ \mathrm{m\,s^{-1}} \bigr)^{2} = 1.5033 \times 10^{-10}\ \mathrm{J}, \] \[ K = 0.0607520 \times 1.5033 \times 10^{-10}\ \mathrm{J} = 9.133 \times 10^{-12}\ \mathrm{J} = 57.0\ \mathrm{MeV} . \]

Units: \( \mathrm{kg} \cdot \mathrm{m^2\,s^{-2}} = \mathrm{J} \), and \( \gamma - 1 \) is dimensionless, as it must be since it is a power series in the dimensionless \( \beta^2 \). The conversion uses \( 1\ \mathrm{MeV} = 1.602176634 \times 10^{-13}\ \mathrm{J} \), exact because the elementary charge is fixed in SI; \( m_p c^2 = 938.3\ \mathrm{MeV} \), the familiar proton rest energy, is a useful check that no factor of \( c \) has been dropped. A

\[ R_u = 1 \ \text{ in } u = \beta^2 \quad \Longleftrightarrow \quad |v| \lt c, \qquad K(1.00 \times 10^{8}\ \mathrm{m\,s^{-1}}) = 9.13 \times 10^{-12}\ \mathrm{J} = 57.0\ \mathrm{MeV}. \]

Reading. Cauchy–Hadamard converts a statement about binomial coefficients into a statement about physics: the series representation of \( \gamma \) is valid for every speed below \( c \) and for no speed above it, and the radius is set by the singularity of \( (1-u)^{-1/2} \) at \( u = 1 \), that is, at \( v = c \). The convergence rate at a given speed is governed by \( \beta^2 \), so at \( \beta \approx 1/3 \) five terms suffice while at \( \beta = 0.99 \) hundreds do not.

Scope. The boundary case \( u = 1 \) (\( v = c \)) is outside the theorem, and here the series genuinely diverges, since \( a_n \ge 1/(2n+1) \) gives \( \sum_n a_n \ge \sum_n 1/(2n+1) = \infty \) — consistent with \( \gamma \to \infty \). The low-speed truncation \( K \approx \tfrac12 m_p v^2 \) is the \( n = 1 \) term, which here would give \( 8.36 \times 10^{-12}\ \mathrm{J} \), an \( 8.4\% \) underestimate.

Problems

Find the radius and centre of the disc of convergence of \( \displaystyle\sum_{n=1}^{\infty} \frac{n^3}{3^n} (z+1)^n \), and decide the behaviour at every point of the boundary circle.
Solution

The centre is \( z_0 = -1 \) and \( c_n = n^3 / 3^n \). Then \[ |c_n|^{1/n} = \frac{(n^3)^{1/n}}{3} = \frac{1}{3} \exp\!\left( \frac{3 \ln n}{n} \right) \longrightarrow \frac{1}{3} \cdot e^{0} = \frac{1}{3}, \] since \( \ln n / n \to 0 \) and \( \exp \) is continuous. The limit exists, so \( \alpha = 1/3 \) and \[ R = \frac{1}{1/3} = 3 . \] The series converges absolutely on \( |z + 1| \lt 3 \) — on the real line, the interval \( -4 \lt z \lt 2 \) — and diverges on \( |z+1| \gt 3 \).

Boundary: if \( |z + 1| = 3 \) then \( |c_n (z+1)^n| = \frac{n^3}{3^n} \cdot 3^n = n^3 \to \infty \), so the terms do not tend to \( 0 \) and the series diverges at every point of the circle. For instance at \( z = 2 \) the terms are \( 1, 8, 27, 64, \dots \). Here the boundary behaviour happens to be uniform, but that is a fact established by hand, not by the theorem.
Find the radius of convergence of \( \displaystyle\sum_{n=0}^{\infty} \frac{(3n)!}{(n!)^3} z^n \) and state the disc numerically to four decimal places.
Solution

All coefficients are positive, so the ratio corollary applies if its limit exists. Compute \[ \frac{c_{n+1}}{c_n} = \frac{(3n+3)!}{((n+1)!)^3} \cdot \frac{(n!)^3}{(3n)!} = \frac{(3n+1)(3n+2)(3n+3)}{(n+1)^3} . \] Divide numerator and denominator by \( n^3 \): \[ \frac{(3 + \frac{1}{n})(3 + \frac{2}{n})(3 + \frac{3}{n})}{(1 + \frac{1}{n})^3} \longrightarrow \frac{3 \cdot 3 \cdot 3}{1} = 27 . \] By the ratio corollary, \( \limsup_n |c_n|^{1/n} = 27 \) and \[ R = \frac{1}{27} = 0.0370\ldots \] So the series converges absolutely for \( |z| \lt 1/27 \approx 0.0370 \) and diverges for \( |z| \gt 1/27 \). Sanity check with the first coefficients: \( c_0 = 1 \), \( c_1 = 6 \), \( c_2 = 90 \), \( c_3 = 1680 \), whose successive ratios \( 6, 15, 18.67, \dots \) do climb towards \( 27 \). (The sum is a hypergeometric function; the boundary point \( z = 1/27 \) is exactly where its singularity sits, and there the series diverges because \( c_n 27^{-n} \sim \frac{\sqrt{3}}{2\pi n} \) is not summable — but that last step needs Stirling and is not part of Cauchy–Hadamard.)
Let \( f(z) = \displaystyle\sum_{k=0}^{\infty} 2^{-k} z^{2^k} \). (a) Write down the coefficient sequence \( (c_n) \) and compute \( R \). (b) Explain why no ratio criterion can be applied. (c) Show that the series converges absolutely and uniformly on the closed disc \( |z| \le 1 \).
Solution

(a) \( c_n = 2^{-k} \) if \( n = 2^k \) for some integer \( k \ge 0 \), and \( c_n = 0 \) otherwise. For the nonzero ones, \[ |c_n|^{1/n} = \bigl( 2^{-k} \bigr)^{1/2^k} = 2^{-k/2^k} \longrightarrow 2^{0} = 1, \] because \( k/2^k \to 0 \) (exponentials beat linear growth). Numerically: \( k = 1 \) gives \( 2^{-1/2} = 0.7071 \), \( k = 4 \) gives \( 2^{-4/16} = 0.8409 \), \( k = 10 \) gives \( 2^{-10/1024} = 0.9933 \). All other terms of the sequence \( |c_n|^{1/n} \) are \( 0 \). Every value is at most \( 1 \) (the value \( 1 \) itself occurs once, at \( k = 0 \), where \( n = 1 \) and \( c_1 = 2^{0} = 1 \)), and the nonzero values tend to \( 1 \) — not monotonically, since \( k = 1 \) and \( k = 2 \) both give \( 2^{-1/2} \), but increasingly from \( k = 2 \) on. So every tail supremum \( s_N = \sup_{n \ge N} |c_n|^{1/n} \) equals \( 1 \); hence \( \alpha = \inf_N s_N = 1 \) and \( R = 1 \).

(b) Between consecutive powers of two there are arbitrarily long blocks of zero coefficients, so \( c_{n+1}/c_n \) is undefined (division by zero) for infinitely many \( n \); the hypothesis “\( c_n \neq 0 \) eventually” of the ratio corollary fails irreparably. The root formula is untroubled: zeros contribute \( 0 \) to a \( \limsup \) of nonnegative numbers and are simply ignored.

(c) On \( |z| \le 1 \), \( |2^{-k} z^{2^k}| \le 2^{-k} =: M_k \) and \( \sum_{k \ge 0} M_k = 2 \lt \infty \). By the Weierstrass M-test the series converges absolutely and uniformly on the closed unit disc, so \( f \) is continuous on \( |z| \le 1 \). This is a case where the boundary behaviour is as good as possible — and yet, by the Ostrowski–Hadamard gap theorem (the exponents satisfy \( 2^{k+1}/2^{k} = 2 \gt 1 \)), \( f \) cannot be analytically continued past any point of \( |z| = 1 \). Convergence on the boundary and continuation past the boundary are different things.
Prove that \( \displaystyle\sum_{n=0}^{\infty} c_n z^n \) and its termwise derivative \( \displaystyle\sum_{n=1}^{\infty} n c_n z^{n-1} \) have the same radius of convergence, and give an example showing that their boundary behaviour can differ.
Solution

For \( z \neq 0 \) the series \( \sum_{n \ge 1} n c_n z^{n-1} \) and \( \sum_{n \ge 1} n c_n z^{n} \) converge for exactly the same \( z \) (they differ by the nonzero factor \( z \)), and both are unaffected at \( z = 0 \); so it suffices to show \[ \limsup_{n \to \infty} |n c_n|^{1/n} = \limsup_{n \to \infty} |c_n|^{1/n} = \alpha . \]

Write \( t_n = n^{1/n} \) and \( x_n = |c_n|^{1/n} \), so \( |n c_n|^{1/n} = t_n x_n \). Since \( \ln t_n = \ln n / n \to 0 \), we have \( t_n \to 1 \). Suppose first \( \alpha \lt \infty \) and fix \( \varepsilon \in (0,1) \). There is \( N_0 \) with \( 1 - \varepsilon \lt t_n \lt 1 + \varepsilon \) for \( n \ge N_0 \), hence for \( N \ge N_0 \) \[ (1-\varepsilon) \sup_{n \ge N} x_n \;\le\; \sup_{n \ge N} t_n x_n \;\le\; (1+\varepsilon) \sup_{n \ge N} x_n . \] Letting \( N \to \infty \) gives \( (1-\varepsilon)\alpha \le \limsup_n t_n x_n \le (1+\varepsilon) \alpha \), and since \( \varepsilon \in (0,1) \) was arbitrary, \( \limsup_n t_n x_n = \alpha \). If \( \alpha = +\infty \), the left inequality alone gives \( \limsup_n t_n x_n \ge (1-\varepsilon) \cdot \infty = \infty \). Either way the two upper limits agree, so the radii \( 1/\alpha \) agree.

Boundary example. \( \sum_{n \ge 1} z^n/n^2 \) has \( R = 1 \) and converges absolutely at every point of \( |z| = 1 \) (comparison with \( \sum 1/n^2 = \pi^2/6 \)). Its termwise derivative is \( \sum_{n \ge 1} z^{n-1}/n \), also of radius \( 1 \), but at \( z = 1 \) it is the harmonic series and diverges. Radius is preserved; boundary convergence is not. Differentiating again gives \( \sum_{n \ge 2} (n-1) z^{n-2} / n \), still radius \( 1 \), now divergent at every boundary point since its terms do not tend to \( 0 \).
Show that \( \displaystyle\sum_{n \ge 0} z^n \), \( \displaystyle\sum_{n \ge 1} \frac{z^n}{n} \) and \( \displaystyle\sum_{n \ge 1} \frac{z^n}{n^2} \) all have \( R = 1 \), and classify the behaviour of each at every point of \( |z| = 1 \). Conclude precisely what Cauchy–Hadamard can and cannot decide.
Solution

Radii. For the first, \( |c_n|^{1/n} = 1 \) for all \( n \), so \( \alpha = 1 \). For the second, \( |c_n|^{1/n} = n^{-1/n} = \exp(-\ln n / n) \to 1 \); for the third, \( n^{-2/n} = \exp(-2 \ln n / n) \to 1 \). In all three cases \( \alpha = 1 \) and \( R = 1/1 = 1 \). The three coefficient sequences are indistinguishable at the level of exponential growth.

\( \sum z^n \) on \( |z| = 1 \). \( |z^n| = 1 \) for every \( n \), so the terms do not tend to \( 0 \) and the series diverges at every one of the boundary points. (Its partial sums are \( (1 - z^{N+1})/(1-z) \) for \( z \neq 1 \), which oscillate without converging.)

\( \sum z^n/n^2 \) on \( |z| = 1 \). \( |z^n/n^2| = 1/n^2 \) and \( \sum_{n \ge 1} 1/n^2 = \pi^2/6 \approx 1.6449 \lt \infty \), so the series converges absolutely at every boundary point, and uniformly on \( |z| \le 1 \) by the M-test. At \( z = 1 \) the sum is \( \pi^2/6 \); at \( z = -1 \) it is \( -\pi^2/12 \approx -0.8225 \).

\( \sum z^n/n \) on \( |z| = 1 \). At \( z = 1 \) it is the harmonic series, which diverges. For \( |z| = 1 \) with \( z \neq 1 \), apply Dirichlet's test: the partial sums of \( z^n \) are bounded, \[ \left| \sum_{n=1}^{N} z^n \right| = \left| \frac{z(1 - z^N)}{1 - z} \right| \le \frac{2}{|1-z|}, \] a bound independent of \( N \), while \( 1/n \) decreases monotonically to \( 0 \). Hence the series converges (conditionally, since \( \sum 1/n \) diverges) at every boundary point except \( z = 1 \). For instance at \( z = -1 \) it is \( \sum (-1)^n/n = -\ln 2 \approx -0.693147 \).

Conclusion. The three series share the same \( \alpha \), hence the same \( R \), hence the same open disc of convergence; on the boundary they exhibit divergence everywhere, convergence everywhere except one point, and absolute convergence everywhere. Since the theorem's input is only \( \limsup |c_n|^{1/n} \), it cannot in principle say anything about \( |z| = 1 \); resolving the circle requires tools sensitive to more than exponential growth — comparison, Dirichlet's or Abel's test, and, for the analytic content, Abel's limit theorem and the theory of singularities on the circle of convergence.