Power Series and Radius of Convergence

For which values of xx does the infinite polynomial ∑n=0∞anxn\sum_{n=0}^\infty a_n x^n actually converge? A power series is a formal device — a limit of partial sums. It is not at all obvious a priori that plugging in a number xx should produce a convergent numerical series. Yet power series are the backbone of analysis. They define exe^x, sin⁡x\sin x, log⁡(1+x)\log(1+x), and every function that is analytic. Analytic here means admitting a local Taylor expansion. Understanding exactly where a power series converges is therefore not a technicality. It is a prerequisite for using these objects at all. The remarkable fact — and the centerpiece of this article — is this. The set of xx for which ∑anxn\sum a_n x^n converges is always an interval centered at the origin. This is called the interval of convergence. Its "radius" is computed by a single formula involving lim sup⁡\limsup of the coefficients. This is the Cauchy–Hadamard theorem. It reduces a seemingly delicate convergence question to one number, R∈[0,∞]R \in [0,\infty]. That number can be read off directly from the coefficients ana_n, with no reference to xx at all. The big picture is this: convergence of a power series is governed entirely by the exponential growth rate of its coefficients. This is exactly the quantity the Root Test is built to detect. Inside the disc ∣x∣<R|x| < R the series converges absolutely. This means it converges unconditionally, with term rearrangements permitted. Outside ∣x∣>R|x| > R it diverges so badly that its terms do not even tend to zero. Only on the boundary circle ∣x∣=R|x| = R does the theorem fall silent. This is a set of measure zero in the grand scheme, but the site of all the interesting subtlety. This article proves the Cauchy–Hadamard formula from the Root Test. It explains why the radius is exactly the reciprocal of lim sup⁡∣an∣1/n\limsup |a_n|^{1/n}. It works through the full procedure for finding an interval of convergence, including the delicate boundary analysis. It sets up, without resolving, the finer question of exactly what happens at the two endpoints. That question is the subject of Behavior of Power Series at the Endpoints.

Key takeaways
  • A power series ∑an(x−c)n\sum a_n(x-c)^n converges absolutely on an open disc ∣x−c∣<R|x - c| < R and diverges for ∣x−c∣>R|x-c| > R.
  • The radius is given by the Cauchy–Hadamard formula R=1/lim sup⁡n→∞∣an∣1/nR = 1/\limsup_{n\to\infty} |a_n|^{1/n}, with the conventions 1/0=∞1/0=\infty and 1/∞=01/\infty = 0.
  • The proof is a direct application of the Root Test to the series of absolute values ∣an∣∣x−c∣n|a_n||x-c|^n.
  • On the boundary circle ∣x−c∣=R|x-c|=R the theorem gives no information — convergence, divergence, and conditional convergence are all possible there.
  • The interval of convergence is (c−R,c+R)(c-R,c+R) possibly union one or both endpoints; determining the endpoints requires separate tests.
  • Key idea: the exponential growth rate of ∣an∣|a_n|, captured by lim sup⁡∣an∣1/n\limsup |a_n|^{1/n}, completely determines the open disc of absolute convergence, and nothing more.

Where this lives today

Power series and their radius of convergence are foundational far beyond a single course. In complex analysis, the Cauchy–Hadamard radius is exactly the radius of the largest disc on which a holomorphic function's Taylor series represents it, tying directly into Cauchy's integral formula and the theory of analytic continuation. In differential equations, the method of Frobenius and power-series solutions to linear ODEs (e.g. Legendre's equation, Bessel's equation) relies on radius-of-convergence estimates to guarantee that formal solutions are genuine functions near a point. In combinatorics and computer science, generating functions are formal power series whose radius of convergence (via Cauchy–Hadamard applied to counting sequences) governs the asymptotic growth rate of the sequence being counted — this is the engine behind singularity analysis of generating functions in analytic combinatorics. In numerical analysis and scientific computing, radius-of-convergence estimates tell you how far a Taylor-series-based numerical scheme (e.g. for solving ODEs or evaluating special functions) can be trusted before truncation error explodes. In probability and statistics, moment generating functions and probability generating functions are power series whose radius of convergence determines which moments exist and how sharply tail probabilities decay (via the relationship between exponential growth of moments and radius of convergence). In cryptography and coding theory, formal power series over finite fields (with a purely algebraic, non-Archimedean analogue of these convergence ideas) underlie stream cipher and error-correcting-code constructions.

Ask the tutor: How do you determine, for a specific power series, exactly which of the four boundary-behavior patterns (converge/diverge at each endpoint) occurs?

DefinitionPower Series and Radius of Convergence

Let (an)n=0∞(a_n)_{n=0}^\infty be a sequence of real (or complex) numbers and let cc be a real (or complex) number. A power series centered at cc is a formal series

∑n=0∞an(x−c)n=a0+a1(x−c)+a2(x−c)2+⋯ ,\sum_{n=0}^\infty a_n (x-c)^n = a_0 + a_1(x-c) + a_2(x-c)^2 + \cdots,

regarded as a function of the variable xx. The numbers ana_n are the coefficients. Define

L=lim sup⁡n→∞∣an∣1/n∈[0,∞],L = \limsup_{n\to\infty} |a_n|^{1/n} \in [0,\infty],

using the convention lim sup⁡\limsup of a sequence that is eventually +∞+\infty is +∞+\infty, and interpreting ∣an∣1/n=0|a_n|^{1/n}=0 when an=0a_n = 0. The radius of convergence of the power series is

R={1L,0<L<∞,0,L=∞,∞,L=0.R = \begin{cases} \dfrac{1}{L}, & 0 < L < \infty,\\[4pt] 0, & L = \infty,\\[4pt] \infty, & L = 0. \end{cases}

The interval (disc) of convergence is the set of all xx for which the series converges; by the theorem below this set always contains the open interval (c−R,c+R)(c-R, c+R) (all of R\mathbb{R}, or all of C\mathbb{C}, if R=∞R = \infty), and contains no point with ∣x−c∣>R|x-c| > R.

Intuition

Think of the coefficients ana_n as measuring how "fast" the series wants to grow. Each term is an(x−c)na_n(x-c)^n, a product of two competing effects: ∣an∣|a_n|, which may grow or shrink with nn, and ∣x−c∣n|x-c|^n, which grows geometrically whenever ∣x−c∣>1|x-c|>1 and shrinks geometrically whenever ∣x−c∣<1|x-c|<1. The question of convergence is a tug-of-war between these two exponential rates. The Root Test resolves any such tug-of-war by looking at the nn-th root of the nn-th term, which strips away the polynomial-order fluctuations in ana_n and leaves exactly the exponential rate ∣an∣1/n⋅∣x−c∣|a_n|^{1/n}\cdot|x-c|. Convergence happens exactly when this quantity's limsup is less than 11, i.e. exactly when ∣x−c∣<1/lim sup⁡∣an∣1/n|x-c| < 1/\limsup|a_n|^{1/n}. Geometrically, this means the set of convergence is a disc (in C\mathbb{C}) or an interval (in R\mathbb{R}), always centered at cc, whose radius RR is a single nonnegative extended real number computed once and for all from the coefficients — it does not depend on which direction from cc you approach. The boundary of that disc, where ∣x−c∣=R|x - c| = R exactly, is where the growth rates are perfectly balanced in the limsup sense, and the outcome can go either way term-by-term; no universal formula can decide it, which is why it demands separate case analysis.

Figure
The three convergence zones on the real line, radius RR centered at cc
c−R c c+R absolute convergence divergence divergence endpoints: outcome not determined by R
TheoremCauchy–Hadamard

Let ∑n=0∞an(x−c)n\sum_{n=0}^\infty a_n(x-c)^n be a power series, and let

L=lim sup⁡n→∞∣an∣1/n,R=1/LL = \limsup_{n\to\infty} |a_n|^{1/n}, \qquad R = 1/L

with the conventions 1/0=∞1/0=\infty, 1/∞=01/\infty=0 as above. Then: 1. If ∣x−c∣<R|x - c| < R, the series converges absolutely. 2. If ∣x−c∣>R|x-c| > R, the terms an(x−c)na_n(x-c)^n do not tend to 00, so the series diverges. 3. If R=0R = 0, the series converges only at x=cx = c. 4. If R=∞R = \infty, the series converges absolutely for every xx. Moreover RR is the unique extended real number in [0,∞][0,\infty] for which (1) and (2) hold; that is, RR is uniquely determined as the supremum of { r≥0:∑anrn converges }\{\,r \ge 0 : \sum a_n r^n \text{ converges}\,\}.

Proof

We work with x−cx - c replaced by a single variable t=x−ct = x - c; proving the statements for ∑antn\sum a_n t^n as a function of tt is equivalent to the original statements. We use the Root Test: for a series ∑bn\sum b_n, if ρ=lim sup⁡n∣bn∣1/n\rho = \limsup_n |b_n|^{1/n}, then ∑bn\sum b_n converges absolutely if ρ<1\rho < 1 and diverges (terms do not tend to 00) if ρ>1\rho > 1.

Step 1 (compute ρ\rho for our series). Apply the root test to bn=antnb_n = a_n t^n. For t≠0t \neq 0,

∣bn∣1/n=∣an∣1/n ∣t∣.|b_n|^{1/n} = |a_n|^{1/n}\,|t|.

By the standard limsup algebra rule for multiplying a sequence by a fixed positive constant (which follows directly from the ε\varepsilon-definition of Limit Superior and Limit Inferior: for λ>0\lambda > 0, lim sup⁡(λcn)=λlim sup⁡cn\limsup(\lambda c_n) = \lambda \limsup c_n), we get

ρ:=lim sup⁡n→∞∣antn∣1/n=∣t∣⋅lim sup⁡n→∞∣an∣1/n=∣t∣⋅L.\rho := \limsup_{n\to\infty} |a_n t^n|^{1/n} = |t|\cdot \limsup_{n\to\infty} |a_n|^{1/n} = |t| \cdot L.

For t=0t = 0 every term with n≥1n \ge 1 vanishes, so the series trivially converges (to a0a_0); we treat t=0t=0 separately below and it is consistent with every case.

Step 2 (case 0<L<∞0 < L < \infty, so R=1/LR = 1/L). For t≠0t \neq 0, ρ=∣t∣L\rho = |t| L.

  • If ∣t∣<R=1/L|t| < R = 1/L, then ρ=∣t∣L<1\rho = |t|L < 1. By the Root Test, ∑antn\sum a_n t^n converges absolutely. This proves (1).
  • If ∣t∣>R=1/L|t| > R = 1/L, then ρ=∣t∣L>1\rho = |t|L > 1. By the Root Test, the terms antna_n t^n do not tend to 00 (indeed the root test's proof shows ∣antn∣|a_n t^n| is eventually ≥\ge some fixed number greater than a term of a divergent geometric-type comparison; concretely: since lim sup⁡∣antn∣1/n>1\limsup |a_n t^n|^{1/n} > 1, there are infinitely many nn with ∣antn∣1/n>1|a_n t^n|^{1/n} > 1, i.e. ∣antn∣>1|a_n t^n| > 1 infinitely often, so antn↛0a_nt^n \not\to 0). Since the general term does not tend to 00, the series ∑antn\sum a_n t^n diverges by the Divergence Test (a series with terms not tending to zero cannot converge — this is immediate from the Cauchy criterion for series, requiring antn→0a_n t^n \to 0 as a necessary condition). This proves (2).

Step 3 (case L=∞L = \infty, so R=0R = 0). Statement (3) asserts convergence only at t=0t = 0. At t=0t=0 the series is just a0a_0, which converges trivially. For any t≠0t \neq 0: since L=∞L = \infty, for every M>0M > 0 there are infinitely many nn with ∣an∣1/n>M|a_n|^{1/n} > M. Choose M=2/∣t∣M = 2/|t|. Then infinitely often ∣an∣1/n>2/∣t∣|a_n|^{1/n} > 2/|t|, i.e. ∣antn∣1/n>2|a_n t^n|^{1/n} > 2, i.e. ∣antn∣>2n→∞|a_n t^n| > 2^n \to \infty. So antn↛0a_n t^n \not\to 0 along that subsequence, hence the series diverges by the same divergence-test argument as in Step 2. This proves (3), matching the formula R=0R=0.

Step 4 (case L=0L = 0, so R=∞R = \infty). Statement (4) asserts absolute convergence for every tt. Fix any tt. If t=0t = 0 this is trivial. If t≠0t \neq 0: since L=0L = 0 , we have ρ=∣t∣⋅0=0\rho = |t|\cdot 0 = 0 by Step 1's computation (limsup of ∣an∣1/n|a_n|^{1/n} is 00 means for every ε>0\varepsilon>0, eventually ∣an∣1/n<ε|a_n|^{1/n}<\varepsilon; taking ε=1/(2∣t∣)\varepsilon = 1/(2|t|) gives eventually ∣antn∣1/n<1/2|a_nt^n|^{1/n} < 1/2, so ρ≤1/2<1\rho \le 1/2 < 1). By the Root Test, the series converges absolutely. This proves (4), matching R=∞R = \infty.

Step 5 (uniqueness of RR). Suppose R′R' also satisfies: the series converges absolutely for ∣t∣<R′|t| < R' and antn↛0a_nt^n \not\to 0 (hence diverges) for ∣t∣>R′|t| > R'. We show R′=RR' = R. If R′<RR' < R, pick tt with R′<∣t∣<RR' < |t| < R (possible since R′<RR' < R; if R=∞R = \infty any ∣t∣>R′|t| > R' works, if R<∞R < \infty pick ∣t∣|t| strictly between). By property (1) already proved for RR, since ∣t∣<R|t| < R the series converges absolutely at tt. But ∣t∣>R′|t| > R' forces divergence by hypothesis on R′R' — contradiction, since absolute convergence implies convergence. If R′>RR' > R, pick tt with R<∣t∣<R′R < |t| < R' (symmetric argument, using R<∞R < \infty; if R=∞R=\infty this case is vacuous since no R′>∞R' > \infty exists in [0,∞][0,\infty]). By property (2) already proved for RR, since ∣t∣>R|t|>R the terms antn↛0a_nt^n \not\to 0, so the series diverges. But ∣t∣<R′|t| < R' forces convergence (indeed absolute convergence) by the hypothesis on R′R' — contradiction. Hence R′=RR' = R, proving uniqueness. Combined with Steps 2–4, R=1/LR = 1/L is exactly the radius separating absolute convergence from term-divergence, completing the proof of the theorem. ■\blacksquare

∎
CorollaryInterval of convergence structure

Let R∈[0,∞]R \in [0,\infty] be the radius of convergence of ∑an(x−c)n\sum a_n(x-c)^n. Then the set of x∈Rx \in \mathbb{R} for which the series converges is exactly one of:

{c} (if R=0),(c−R,c+R), [c−R,c+R), (c−R,c+R], or [c−R,c+R] (if 0<R<∞),R (if R=∞).\{c\}\ (\text{if } R=0), \qquad (c-R,c+R),\ [c-R,c+R),\ (c-R,c+R],\ \text{or } [c-R,c+R]\ (\text{if }0<R<\infty), \qquad \mathbb{R}\ (\text{if }R=\infty).

In particular this set is always an interval centered at cc (with cc itself as midpoint when RR is finite and positive), and it is not possible to further shrink or enlarge the open middle portion (c−R,c+R)(c-R,c+R): this open interval is contained in the convergence set and its complement's interior is contained in the divergence set, by parts (1) and (2) of the theorem.

Proof

By part (1) of the theorem every point of (c−R,c+R)(c-R, c+R) lies in the convergence set (in fact the absolute-convergence set), and by part (2) every point outside [c−R,c+R][c-R,c+R] lies outside the convergence set. So the convergence set differs from (c−R,c+R)(c-R,c+R) only possibly by inclusion of the two endpoints c−Rc - R and c+Rc+R individually. This gives exactly the four listed possibilities when 0<R<∞0 < R < \infty; the two degenerate cases R=0R = 0 and R=∞R = \infty follow directly from parts (3) and (4). ■\blacksquare

∎

Worked examples

Example 1 — an asymmetric coefficient pattern requiring careful limsup computation. Consider the power series centered at c=−1c = -1:

∑n=1∞an(x+1)n,an={3nn,n even2nn2,n odd.\sum_{n=1}^\infty a_n (x+1)^n, \qquad a_n = \begin{cases} \dfrac{3^n}{n}, & n \text{ even} \\[4pt] \dfrac{2^n}{n^2}, & n \text{ odd.} \end{cases}

We find the radius and interval of convergence. Step 1: compute ∣an∣1/n|a_n|^{1/n} along each subsequence. For nn even,

∣an∣1/n=(3nn)1/n=3n1/n→3,|a_n|^{1/n} = \left(\frac{3^n}{n}\right)^{1/n} = \frac{3}{n^{1/n}} \to 3,

using the standard fact n1/n→1n^{1/n} \to 1 (which follows from n1/n=e(ln⁡n)/n→e0=1n^{1/n} = e^{(\ln n)/n} \to e^0 = 1). For nn odd,

∣an∣1/n=(2nn2)1/n=2n2/n→2,|a_n|^{1/n} = \left(\frac{2^n}{n^2}\right)^{1/n} = \frac{2}{n^{2/n}} \to 2,

since n2/n=(n1/n)2→1n^{2/n} = (n^{1/n})^2 \to 1. Step 2: compute the limsup. The sequence ∣an∣1/n|a_n|^{1/n} has one subsequence (even nn) converging to 33 and another (odd nn) converging to 22. The set of subsequential limits is {2,3}\{2,3\}, so

L=lim sup⁡n→∞∣an∣1/n=max⁡{2,3}=3.L = \limsup_{n\to\infty}|a_n|^{1/n} = \max\{2,3\} = 3.

(This uses the characterization of lim sup⁡\limsup as the largest subsequential limit, justified in Limit Superior and Limit Inferior.) Step 3: radius. R=1/L=1/3R = 1/L = 1/3. Step 4: interval of convergence before checking endpoints. By Cauchy–Hadamard, the series converges absolutely for ∣x−(−1)∣<1/3|x - (-1)| < 1/3, i.e.

−43<x<−23,-\tfrac{4}{3} < x < -\tfrac{2}{3},

and diverges for ∣x+1∣>1/3|x+1| > 1/3. Step 5: check the endpoints x=−4/3x = -4/3 and x=−2/3x=-2/3 directly (they are not covered by the theorem). At x=−2/3x = -2/3: x+1=1/3x + 1 = 1/3, so the general term is an(1/3)na_n(1/3)^n. For even nn: an(1/3)n=3nn⋅13n=1na_n(1/3)^n = \frac{3^n}{n}\cdot\frac{1}{3^n} = \frac1n. For odd nn: an(1/3)n=2nn2⋅13n=1n2(23)n→0a_n(1/3)^n = \frac{2^n}{n^2}\cdot\frac{1}{3^n} = \frac{1}{n^2}\left(\frac23\right)^n \to 0 rapidly (geometric decay). So the series splits as

∑n even1n+∑n odd1n2(23)n.\sum_{n \text{ even}} \frac{1}{n} + \sum_{n\text{ odd}} \frac{1}{n^2}\left(\frac23\right)^n.

The odd part converges absolutely (compare to a convergent geometric series, ratio 2/3<12/3<1, dominating polynomial factor). The even part is (half of) the harmonic series, which diverges. A convergent series plus a divergent series is divergent (if the sum converged, subtracting the convergent piece would force the divergent piece to converge — contradiction). So the full series diverges at x=−2/3x=-2/3. At x=−4/3x = -4/3: x+1=−1/3x+1 = -1/3, so the general term is an(−1/3)na_n(-1/3)^n. For even nn, (−1/3)n=(1/3)n(-1/3)^n = (1/3)^n (since nn even), giving again 1n\frac{1}{n}, harmonic and divergent, with no sign alternation to help (all these terms are positive since nn is even). For odd nn, (−1/3)n=−(1/3)n(-1/3)^n = -(1/3)^n, giving −1n2(2/3)n→0-\frac{1}{n^2}(2/3)^n \to 0 absolutely convergent. Again convergent-plus-divergent is divergent, so the series diverges at x=−4/3x = -4/3 too. Conclusion: the interval of convergence is the open interval (−43,−23)\left(-\frac43, -\frac23\right), radius R=1/3R = 1/3, with divergence at both endpoints.

Example 2 — a series with factorial-type coefficients and a shifted, non-integer center. Consider

∑n=0∞(−1)n(n!)2(2n)! (x−5)n.\sum_{n=0}^\infty \frac{(-1)^n (n!)^2}{(2n)!}\,(x - 5)^n.

We determine the radius of convergence, this time using the ratio form of Cauchy–Hadamard (valid when the limit of ratios exists, via the standard fact that lim⁡∣an+1/an∣\lim |a_{n+1}/a_n|, if it exists, equals lim⁡∣an∣1/n=lim sup⁡∣an∣1/n\lim|a_n|^{1/n} = \limsup |a_n|^{1/n} — a consequence of the Root Test's comparison with the Ratio Test). Step 1: set up the ratio. Let an=(−1)n(n!)2(2n)!a_n = \dfrac{(-1)^n(n!)^2}{(2n)!}, so ∣an∣=(n!)2(2n)!|a_n| = \dfrac{(n!)^2}{(2n)!}. Then

∣an+1∣∣an∣=((n+1)!)2(2n+2)!⋅(2n)!(n!)2=(n+1)2(n!)2(2n+2)(2n+1)(2n)!⋅(2n)!(n!)2=(n+1)2(2n+2)(2n+1).\frac{|a_{n+1}|}{|a_n|} = \frac{((n+1)!)^2}{(2n+2)!}\cdot \frac{(2n)!}{(n!)^2} = \frac{(n+1)^2 (n!)^2}{(2n+2)(2n+1)(2n)!}\cdot\frac{(2n)!}{(n!)^2} = \frac{(n+1)^2}{(2n+2)(2n+1)}.

Step 2: simplify and take the limit. Note (2n+2)=2(n+1)(2n+2) = 2(n+1), so

(n+1)22(n+1)(2n+1)=n+12(2n+1).\frac{(n+1)^2}{2(n+1)(2n+1)} = \frac{n+1}{2(2n+1)}.

As n→∞n \to \infty,

n+12(2n+1)=1+1/n2(2+1/n)⟶14.\frac{n+1}{2(2n+1)} = \frac{1 + 1/n}{2(2 + 1/n)} \longrightarrow \frac{1}{4}.

Step 3: apply the ratio-limit fact. Since lim⁡n→∞∣an+1∣/∣an∣=1/4\lim_{n\to\infty} |a_{n+1}|/|a_n| = 1/4 exists, it equals lim sup⁡n∣an∣1/n=L\limsup_n |a_n|^{1/n} = L. So L=1/4L = 1/4 and

R=1/L=4.R = 1/L = 4.

Step 4: interval before endpoints. The series converges absolutely for ∣x−5∣<4|x - 5| < 4, i.e. 1<x<91 < x < 9, and diverges for ∣x−5∣>4|x-5|>4, i.e. x<1x < 1 or x>9x > 9. Step 5 (brief endpoint remark, foreshadowing the next topic): at x=9x = 9, x−5=4x - 5 = 4, and the general term is (−1)n(n!)2(2n)!4n\frac{(-1)^n(n!)^2}{(2n)!}4^n; by Stirling-type estimates (n!)24n(2n)!∼πn\frac{(n!)^2 4^n}{(2n)!} \sim \sqrt{\pi n} (using (2nn)∼4n/πn\binom{2n}{n} \sim 4^n/\sqrt{\pi n}), which grows without bound, so the terms do not tend to 00 and the series diverges at x=9x=9. At x=1x=1, x−5=−4x-5=-4, and by the same asymptotic estimate with an extra alternating sign (−1)n⋅(−1)n=1(-1)^n\cdot(-1)^n = 1 (since (−4)n=(−1)n4n(-4)^n = (-1)^n4^n, contributing an extra (−1)n(-1)^n that cancels the existing (−1)n(-1)^n in ana_n), the terms again fail to tend to 00, so the series diverges at x=1x=1 as well. Thus here the interval of convergence is again the fully open interval (1,9)(1,9) — but as the next example category shows, this is not the only possible outcome, which is exactly why endpoint behavior needs its own dedicated analysis (see Behavior of Power Series at the Endpoints).

CounterexampleWhy the limsup — not an ordinary limit — is essential

One might hope to simplify the Cauchy–Hadamard formula by replacing lim sup⁡n∣an∣1/n\limsup_n |a_n|^{1/n} with an ordinary limit lim⁡n∣an∣1/n\lim_n |a_n|^{1/n}, reasoning that "the growth rate should stabilize." This hope fails: for many coefficient sequences, lim⁡n∣an∣1/n\lim_n |a_n|^{1/n} simply does not exist, while lim sup⁡\limsup always exists (as an extended real number) by definition. Dropping the limsup and insisting on an ordinary limit breaks the theorem's applicability entirely for such series. Take the coefficients

an={2−n,n even4−n,n odd,a_n = \begin{cases} 2^{-n}, & n \text{ even} \\ 4^{-n}, & n \text{ odd}, \end{cases}

and consider ∑anxn\sum a_n x^n centered at c=0c=0. Then ∣an∣1/n=1/2|a_n|^{1/n} = 1/2 for even nn and ∣an∣1/n=1/4|a_n|^{1/n} = 1/4 for odd nn. The sequence ∣an∣1/n|a_n|^{1/n} oscillates forever between 1/21/2 and 1/41/4; it has two distinct subsequential limits, so lim⁡n∣an∣1/n\lim_n |a_n|^{1/n} fails to exist as an ordinary limit — there is no single number the sequence "settles down to." If we tried to state Cauchy–Hadamard using an ordinary limit "L=lim⁡∣an∣1/nL = \lim |a_n|^{1/n}, R=1/LR = 1/L," the formula would simply be undefined here, giving no answer at all, even though the radius of convergence plainly exists and is computable: since lim sup⁡∣an∣1/n=max⁡{1/2,1/4}=1/2\limsup |a_n|^{1/n} = \max\{1/2,1/4\} = 1/2, the correct radius is R=1/(1/2)=2R = 1/(1/2) = 2. Indeed one checks directly that ∑anxn\sum a_n x^n converges absolutely for ∣x∣<2|x|<2 (each of the even and odd sub-series is a convergent geometric-type series there) and that the terms fail to tend to 00 for ∣x∣>2|x|>2 along the even-indexed subsequence, confirming R=2R=2 independently. This shows precisely what the ordinary-limit hypothesis would cost: replacing lim sup⁡\limsup by lim⁡\lim in the statement of the theorem is a genuine weakening of the tool (it requires an extra hypothesis — existence of the ordinary limit — that most coefficient sequences do not satisfy), and the conclusion ("R=1/LR=1/L computes the radius") becomes simply inapplicable to a large and completely ordinary class of power series.

Scope, edge cases & common traps

SituationWhat Cauchy–Hadamard tells youWhat it does not tell you
0<R<∞0 < R < \inftyAbsolute convergence on (c−R,c+R)(c-R,c+R); divergence outside [c−R,c+R][c-R,c+R]Behavior exactly at x=c±Rx = c\pm R
R=0R = 0Convergence only at x=cx=cNothing further needed — set is a single point
R=∞R = \inftyAbsolute convergence everywhereNothing further needed — set is all of R\mathbb R
lim⁡∣an∣1/n\lim|a_n|^{1/n} existsLL equals that limit; use it directly—
lim⁡∣an∣1/n\lim|a_n|^{1/n} does not existMust use lim sup⁡\limsup, not an ordinary limitNaive "1/lim⁡1/\lim" formulas are meaningless here
Ratio ∣an+1/an∣→ℓ|a_{n+1}/a_n| \to \elllim sup⁡∣an∣1/n=ℓ\limsup|a_n|^{1/n}=\ell too, so R=1/ℓR=1/\ellThis shortcut requires the ratio limit to exist

The mathematician behind it

Origins · 1 minJacques Hadamard and how this was discovered
JH
Jacques Hadamard
1865–1963 · French analysis and number theory, École Polytechnique and Collège de France

Jacques Hadamard was one of the most versatile analysts of his generation, celebrated for proving the Prime Number Theorem (independently of Charles de la Vallée Poussin, in 1896) using deep properties of the Riemann zeta function, and for foundational contributions to the theory of partial differential equations, functional analysis, and the calculus of variations. His 1892 formula for the radius of convergence of a power series in terms of lim sup⁡∣an∣1/n\limsup |a_n|^{1/n} sharpened and completed an earlier, less general observation attributed to Augustin-Louis Cauchy, giving the theorem its now-standard hyphenated name. Hadamard lived to 98, surviving both World Wars, and mentored a generation of French mathematicians while maintaining an extraordinarily broad research output across analysis, number theory, and mathematical psychology.

History
- 1821 — Cauchy, in his Cours d'Analyse, studies convergence of series systematically and effectively identifies the growth-rate criterion in restricted form.
- 1892 — Hadamard's thesis and related papers establish the general formula R=1/lim sup⁡n∣an∣1/nR = 1/\limsup_n |a_n|^{1/n} for arbitrary complex coefficient sequences, resolving cases Cauchy's more limited statement did not cover.
- 1893–1900s — the formula becomes standard in the developing rigorous theory of complex power series and analytic functions, alongside Weierstrass's program of building analysis on power series.
- 1920s–1930s — the theorem is absorbed into the standard undergraduate and graduate real- and complex-analysis curriculum (Titchmarsh, Whittaker–Watson, and later Rudin all present it essentially as stated here).
- Present — Cauchy–Hadamard remains the universal starting point for radius-of-convergence computations in real analysis, complex analysis, and formal power series over normed fields.
Does the radius of convergence depend on which point cc we center the series at?

Yes and no. The value of RR depends only on the coefficients ana_n, via R=1/lim sup⁡∣an∣1/nR = 1/\limsup|a_n|^{1/n}, and is completely independent of cc. But the disc itself, {x:∣x−c∣<R}\{x : |x-c|<R\}, is centered at whatever cc you chose, so shifting the center moves the interval of convergence rigidly without changing its length. If you re-expand the same function as a Taylor series around a different point c′c', the coefficients typically change (they become derivatives of the function at c′c' divided by factorials), so the new radius can genuinely differ — often it equals the distance from c′c' to the nearest singularity of the function being represented.

Can the radius of convergence be computed directly from the ratio test instead of the root test?

Often, yes, and it is frequently more convenient. If lim⁡n→∞∣an+1/an∣\lim_{n\to\infty}|a_{n+1}/a_n| exists and equals ℓ\ell, then it is a standard fact that lim⁡n∣an∣1/n\lim_n|a_n|^{1/n} also exists and equals ℓ\ell, so lim sup⁡∣an∣1/n=ℓ\limsup|a_n|^{1/n}=\ell too, giving R=1/ℓR=1/\ell directly. This shortcut is exactly what powers Worked Example 2 above. However, the ratio limit does not always exist (e.g. Example 1's coefficients, or the counterexample block), while the limsup in Cauchy–Hadamard always exists as an extended real number — so the root-based formula is the universally valid one, and the ratio-based shortcut is a convenient special case, not a replacement.

What happens right at the boundary ∣x−c∣=R|x - c| = R?

The Cauchy–Hadamard theorem gives absolutely no information there, by design: the Root Test's own proof breaks down exactly when the limsup being tested equals 11, which is precisely what happens on the boundary circle. Convergence there is governed by finer structure of the coefficients — decay rate, sign patterns, and cancellation — rather than by the coarse exponential growth rate that RR captures. All four combinations (converge at both endpoints, diverge at both, converge at exactly one) are genuinely possible, as illustrated by the two worked examples and the standard trio ∑xn\sum x^n, ∑xn/n\sum x^n/n, ∑xn/n2\sum x^n/n^2. This finer analysis is the subject of Behavior of Power Series at the Endpoints.

Is the interval of convergence always symmetric around the center, even when the coefficients look asymmetric (e.g. depend on parity of nn)?

Yes, always — this is a direct and perhaps surprising consequence of the theorem. Even though the coefficients ana_n in Worked Example 1 behave very differently on even versus odd indices, the set of xx for which the series converges absolutely is still forced to be a disc centered exactly at cc, because the criterion ∣t∣⋅L<1|t|\cdot L < 1 depends on xx only through ∣x−c∣|x - c|, never through the sign or direction of x−cx - c. Coefficient asymmetry can only affect the value of LL (hence of RR) and can only affect what happens at the two individual endpoints — it can never make the convergence region itself lopsided in the interior.

Does absolute convergence inside (c−R,c+R)(c-R, c+R) mean the series can be freely rearranged and manipulated there?

Yes. Since the theorem establishes absolute, not merely conditional, convergence for every xx strictly inside the radius, all the standard tools that require Absolute vs Conditional Convergence apply freely there: terms may be rearranged without changing the sum, the series may be added or multiplied term-by-term with another absolutely convergent power series (Cauchy product), and — with further work beyond this article — the series may be differentiated and integrated term by term within the open interval of convergence. None of these manipulations are automatically justified at boundary points, where convergence, if it occurs at all, may only be conditional.

If two power series centered at the same point agree on some interval around the center, must their coefficients be identical?

Within the common region of convergence, yes, assuming both series actually converge to the same function on an interval containing cc. This is the identity theorem for power series, and it follows from the fact that a power series's coefficients are recoverable as an=f(n)(c)/n!a_n = f^{(n)}(c)/n! from the function ff it converges to (differentiating term by term inside the radius of convergence, which is justified once term-by-term differentiability is established). This uniqueness is what makes "the" power series of a function well-defined, and it underlies why Taylor series are unique when they exist, though this article's Cauchy–Hadamard theorem itself only concerns where a given series converges, not this further uniqueness statement.

Why does the Root Test, rather than the Ratio Test, give the fully general version of this theorem?

Because the Root Test's criterion lim sup⁡n∣bn∣1/n\limsup_n |b_n|^{1/n} is defined and finite (or +∞+\infty) for every sequence, with no existence hypothesis needed, since lim sup⁡\limsup of any bounded-below sequence of nonnegative extended reals always exists in [0,∞][0,\infty]. The Ratio Test's criterion lim⁡n∣bn+1/bn∣\lim_n |b_{n+1}/b_n| requires that limit to exist, which fails for many natural coefficient sequences (as in the counterexample above and in Worked Example 1). Since Cauchy–Hadamard is meant to apply to every power series without exception, it must be built on the Root Test; the Ratio Test remains a useful computational shortcut precisely in the common special case where the relevant limit happens to exist.

Before this

Where this leads