Radius of convergence ( $R$ )

The radius of convergence of

\sum_{n \geq 0} a_{n} (z - c)^{n}

is the unique

R \in [0, \infty]

separating interior absolute convergence from exterior divergence. It converges absolutely when

∣ z - c ∣ < R

and diverges when

∣ z - c ∣ > R

. At the centre it always has the finite value

a_{0}

, since all positive powers vanish. The boundary convergence of a power series is a separate question.

For

0 < R < \infty

, the interior is a disk in the complex plane and an open interval

(c - R, c + R)

on the real line. A power series converges uniformly on every smaller closed disk

∣ z - c ∣ \leq r < R

. This gives stable polynomial approximation over a whole compact region, rather than a separate limiting calculation at each point. See Jiří Lebl, Basic Analysis, §11.3.2: convergence of power series.

Four examples reveal what

R

measures:

$\sum_{n \geq 0} z^{n}$ has $R = 1$ , since its coefficients have constant magnitude one.
$\sum_{n \geq 1} z^{n} / n$ also has $R = 1$ : a factor $1/ n$ changes boundary behaviour without changing the limiting coefficient-root growth.
$\sum_{n \geq 0} z^{n} / n!$ has $R = \infty$ . The ratio test gives successive term ratio $∣ z ∣/ (n + 1) \to 0$ for every fixed $z$ , producing the exponential function.
$\sum_{n \geq 0} n! z^{n}$ has $R = 0$ . For any $z \neq = 0$ , successive term magnitudes have ratio $(n + 1) ∣ z ∣ \to \infty$ , so the terms do not tend to zero.

The Cauchy-Hadamard theorem computes

R

directly from the coefficients. The attractive feature is that one growth statistic decides convergence throughout the interior and exterior, including sparse series with many zero coefficients. Finite changes to the coefficients do not alter the radius, though they can change the sum.

Table of contents

Integer coefficients with a prescribed radius of convergence ( $a_{n} = ⌈ R^{- n} ⌉ (0 < R \leq 1)$ )

 0  0

Every

R \in [0, 1]

occurs as the radius of convergence of a power series all of whose coefficients are positive integers. For

0 < R \leq 1

, use the ceiling function to choose

a_{n} = ⌈ R^{- n} ⌉

. Then

R^{- n} \leq a_{n} < R^{- n} + 1 \leq 2 R^{- n} .

(1)

For

∣ z ∣ < R

, the comparison test for series with a geometric series proves absolute convergence. For

∣ z ∣ \geq R

, the bound

∣ a_{n} z^{n} ∣ \geq (∣ z ∣/ R)^{n} \geq 1

violates the fact that terms of a convergent series tend to zero. Hence the radius of convergence is exactly

R

For

R = 0

, use

a_{n} = n!

: at every

z \neq = 0

the ratio of successive term absolute values is

(n + 1) ∣ z ∣

, eventually exceeding two. The factorial coefficients therefore give divergence away from zero.

A power series with every coefficient a positive integer cannot have radius of convergence greater than one:

a_{n} \geq 1

makes convergence impossible at

z = 1

. If zero coefficients are allowed and only finitely many are nonzero, a polynomial instead has infinite radius of convergence.

Disk of convergence ( ${z \in C : ∣ z - c ∣ < R}$ )

 0  0

For a power series centred at

c

with radius of convergence

R

, its disk of convergence is the open disk

∣ z - c ∣ < R

, on which the series has absolute convergence. Boundary points may or may not belong to the larger domain of convergence. When

R = 0

this open disk is empty, although the power series still converges at its centre; when

R = \infty

it is the whole complex plane.

Radius of a Maclaurin series from successive derivatives ( $R = lim_{n \to \infty} (n + 1) f^{(n)} (0) / f^{(n + 1)} (0)$ )

 0  0

Write the Maclaurin series coefficients as

a_{n} = f^{(n)} (0) / n!

. When the successive ratios exist eventually and have a limit, the ratio test gives the displayed radius of convergence, including zero and infinity. The unconditional formula is

R^{- 1} = lim sup_{n \to \infty} ∣ a_{n} ∣^{1/ n}

, from the Cauchy-Hadamard theorem. Endpoints must be checked separately.

The existence and convergence of this power series do not by themselves imply that it equals

f

: equality requires analyticity or control of the Taylor remainder. The smooth function

f (x) = e^{- 1/ x^{2}}

for

x \neq = 0

f (0) = 0

, has all derivatives zero at zero, so its Maclaurin series is identically zero despite the nonzero values away from zero.

Isolating a logarithmic singularity to test a power-series endpoint ( $f (z) = A lo g (1 - z / R) + H (z)$ )

 0  0

Suppose

A \neq = 0

and

R > 0

, and

H

is an analytic function on a disk of radius strictly larger than

R

. The Taylor series at zero of

f (z) = A lo g (1 - z / R) + H (z)

has radius of convergence

R

. Its coefficients are the sum of

- A / (n R^{n})

and the more rapidly decreasing coefficients of

H

. At

x = - R

, the logarithmic part is an alternating harmonic series and converges conditionally; at

x = R

it diverges. The power series for

H

converges absolutely at both endpoints and therefore cannot change either conclusion. The real convergence interval is

[- R, R)

For

f (z) = lo g (2 - e^{z})

, take

R = lo g 2

and

H (z) = lo g [(2 - e^{z}) / (1 - z / R)]

, with its branch normalized by

H (0) = 0

. The quotient has nonzero removable value

2 R

z = R

. Its next zeros are at

R \pm 2 πi

, so a holomorphic logarithm exists on a larger disk. This shows why the Taylor series converges at

x = - lo g 2

even though its singular positive endpoint is excluded.

Boundary convergence of a power series

 0  0

Knowing the radius of convergence does not settle convergence on its boundary. The three power series

\sum_{n = 0}^{\infty} z^{n}, \sum_{n = 1}^{\infty} \frac{z ^{n}}{n}, \sum_{n = 1}^{\infty} \frac{z ^{n}}{n ^{2}}

(1)

all have radius one, but behave differently on

∣ z ∣ = 1

Figure 1.
The same radius of convergence permits three different behaviours on the unit circle
. Green denotes absolute convergence, orange conditional convergence, and red divergence. All three series converge absolutely in the open unit disk. On its boundary, the geometric series diverges everywhere; the $1/ n$ series converges conditionally except at $z = 1$ ; the $1/ n^{2}$ series converges absolutely everywhere. **Details**.

For the geometric series, boundary terms all have complex modulus one, violating the term test for divergence. For the second series, absolute values give the divergent harmonic series at every boundary point. At

z = 1

the series itself is harmonic and diverges. At any other point on the unit circle, its geometric partial sums obey

\sum_{k = 1}^{N} z^{k} = \frac{z ( 1 - z ^{N} )}{1 - z} \leq \frac{2}{∣1 - z ∣} .

(2)

The Dirichlet test with the decreasing sequence

1/ n \to 0

therefore gives conditional convergence. In particular,

z = - 1

gives the negative of the alternating harmonic series, with sum

- ln 2

. For the third series, the convergent p-series

\sum_{n \geq 1} 1/ n^{2}

bounds the absolute values at every boundary point, giving absolute convergence by the comparison test for series.

z = - 1

, the rational function

1/ (1 - z)

is well defined and equals

1/2

, yet its geometric series diverges there. A formula for a function and a particular series representation can have different domains. Likewise, applying termwise differentiation of a power series preserves the radius but need not preserve the boundary convergence.

Bounded power-series terms force interior absolute convergence ( $∣ a_{n} z^{n} ∣ \leq M ⟹ \sum_{n} ∣ a_{n} w^{n} ∣ < \infty (∣ w ∣ < ∣ z ∣)$ )

 0  0

If the terms of a power series at a nonzero point

z

are bounded by

M

, then

∣ a_{n} w^{n} ∣ \leq M (∣ w ∣/∣ z ∣)^{n}

. A convergent geometric series and the comparison test for series give absolute convergence at every point strictly inside that circle. Consequently terms at every point outside the radius of convergence must be unbounded. This conclusion is stronger than merely saying that those terms fail to tend to zero.

Sum of power series with unequal radii of convergence ( $R \neq = S ⟹ R_{sum} = min (R, S)$ )

 0  0

Inside both convergence discs, the triangle inequality gives absolute convergence of the sum. Suppose

R < S

and the sum had radius

T > R

. At any modulus between

R

and

min (T, S)

, subtracting the absolutely convergent second series from the absolutely convergent sum would give absolute convergence of the first series outside its radius of convergence, a contradiction. Interchanging the two series covers

S < R

Terms of a power series are unbounded outside its convergence disc ( $∣ z ∣ > R \Rightarrow sup_{n} ∣ a_{n} z^{n} ∣ = \infty$ )

 0  0

Let a power series

\sum_{n \geq 0} a_{n} z^{n}

have finite radius of convergence

R

. If its terms were bounded in complex modulus at a point

z

with

∣ z ∣ > R

, say

∣ a_{n} z^{n} ∣ \leq M

, choose

w

with

R < ∣ w ∣ < ∣ z ∣

. Then

∣ a_{n} w^{n} ∣ \leq M (\frac{∣ w ∣}{∣ z ∣})^{n} .

(1)

The comparison test for series with a convergent geometric series would give absolute convergence at

w

, contradicting its position outside the convergence disc. Hence the terms are unbounded at every such

z

. The argument also works when

R = 0

and gives a stronger obstruction than the usual term test for divergence.

Radius of convergence of a sparse power series ( $R^{- 1} = lim sup_{n} ∣ a_{n} ∣^{1/ m_{n}}$ )

 0  0

For a series

\sum_{n} a_{n} x^{m_{n}}

with distinct increasing integer exponents, define the ordinary coefficients to be

a_{n}

m_{n}

and zero elsewhere. The Cauchy-Hadamard theorem then gives

R^{- 1} = lim sup_{n} ∣ a_{n} ∣^{1/ m_{n}} .

(1)

The exponent in the root is the actual degree

m_{n}

, not the index

n

. For

a_{n} = (n!)^{2}

and

m_{n} = n^{2}

0 \leq 2 lo g (n!) / n^{2} \leq 2 lo g n / n \to 0

, so

R = 1

. Boundary behavior is a separate question.

Polynomial-coefficient power series ( $\sum_{n \geq 0} p (n) z^{n}$ )

 0  0

For a nonzero polynomial

p

, the power series

\sum_{n \geq 0} p (n) z^{n}

has radius of convergence

1

. If

p

has degree

d

and leading coefficient

c \neq = 0

, then

p (n) \sim c n^{d}

and

∣ p (n + 1) / p (n) ∣ \to 1

, so the ratio test applies inside the unit disk. At every point of its boundary, the terms fail to tend to zero. The zero polynomial is the exceptional case with radius of convergence

\infty

Polynomial coefficient growth with unit power-series radius ( $c \leq a_{n} \leq C (1 + n)^{m} ⟹ R = 1$ )

 0  0

Suppose real coefficients satisfy

c \leq a_{n} \leq C (1 + n)^{m}

for all sufficiently large

n

, where

c, C > 0

and

m \geq 0

. Then the power series

\sum a_{n} z^{n}

has radius of convergence one. For

∣ z ∣ < 1

, the comparison test for series and the ratio test dominate it by a polynomial times a geometric sequence. For

∣ z ∣ \geq 1

, the terms have modulus at least

c

eventually and violate the term test for divergence. Finite changes to the coefficients do not affect the conclusion.

Cauchy-Hadamard theorem ( $R^{- 1} = lim sup_{n \to \infty} ∣ a_{n} ∣^{1/ n}$ )

 0  0

For the power series

\sum_{n \geq 0} a_{n} (z - c)^{n}

, the Cauchy-Hadamard theorem gives

\frac{1}{R} = lim sup_{n \to \infty} ∣ a_{n} ∣^{1/ n},

(1)

with

R = \infty

when the limit superior is zero, and

R = 0

when it is infinite. Away from the centre, the root test explains the formula: the limit superior of the term roots at displacement

z - c

∣ z - c ∣/ R

, interpreted as zero for

R = \infty

and infinite for

R = 0

. Values below one give absolute convergence; values above one imply that the terms fail to tend to zero. A value of one leaves the boundary question open. At the centre the series converges regardless of

R

For

a_{2 n} = 1

and

a_{2 n + 1} = 0

, the coefficient roots alternate between one and zero, so their ordinary limit does not exist. Their limit superior is nevertheless one, giving

R = 1

for

\sum_{n \geq 0} z^{2 n} = 1/ (1 - z^{2})

inside the unit disk. A ratio of consecutive coefficients would be undefined at the zero coefficients; the root-limsup formula handles the gap directly.