Radius of Convergence

One Number Organizes Convergence

A power series can converge at many inputs and diverge at others. Convergence Transfers Inward, proved in “What Is a Power Series?”, shows that convergence at one point away from the center guarantees absolute convergence at every strictly closer point. This suggests that the inputs of convergence are organized by their distance from the center: closer inputs are controlled together, even though two inputs on opposite sides need not have identical behavior.

The radius of convergence captures this structure in a single number. Inside the radius, the series converges absolutely; outside it, the series diverges. The radius does not, by itself, decide what happens exactly at the boundary. We will first define the radius precisely and prove its basic properties, then obtain a formula for calculating it from the growth of the coefficients.

Defining the Radius

For a fixed distance $r\geq0$, the absolute values of the terms of a power series at any point with $|x-a|=r$ are $|c_n|r^n$. Thus, whether the series converges absolutely at that distance depends only on $r$, not on which side of the center the input lies. At $r=0$, interpret the terms as $|c_0|$ for $n=0$ and $0$ for every $n\geq1$.

Definition: For a power series $\sum_{n=0}^{\infty}c_n(x-a)^n$, let $D$ be the set of distances $r\geq0$ for which the numerical series $\sum_{n=0}^{\infty}|c_n|r^n$ converges. The radius of convergence is $R=\sup D$, with $R=\infty$ if $D$ is unbounded. In particular, $0\in D$, since the series at the center has only one possibly nonzero term.

This definition uses absolute convergence, but it also controls ordinary convergence beyond the radius. The key fact is that if a distance belongs to $D$, then every smaller distance does too: the terms at a smaller distance are bounded by the corresponding terms at the larger one. The Comparison Principle for Positive-Term Series justifies this comparison.

Theorem (Convergence Inside and Outside the Radius): Let $R$ be the radius of convergence of $\sum_{n=0}^{\infty}c_n(x-a)^n$. If $|x-a|<R$, the series converges absolutely at $x$. If $|x-a|>R$, the series diverges at $x$. When $R=0$, the first claim applies only at the center; when $R=\infty$, there are no real inputs satisfying the second inequality.

Proof. Write $r=|x-a|$. First suppose $r<R$. By the definition of the supremum, there is some $s\in D$ with $s>r$. Indeed, if there were no such $s$, then $r$ would be an upper bound for $D$, contradicting $r<R=\sup D$. For each $n\geq0$, since $0\leq r<s$,

|c_n|r^n\leq |c_n|s^n.

The series $\sum_{n=0}^{\infty}|c_n|s^n$ converges because $s\in D$. The Comparison Principle for Positive-Term Series therefore gives convergence of $\sum_{n=0}^{\infty}|c_n|r^n$. This is absolute convergence at $x$. If $r=0$, the same conclusion follows directly from convergence of the single possibly nonzero term $|c_0|$.

Now suppose $R<r=|x-a|$. This case can occur only when $R$ is finite. Assume, for contradiction, that the power series converges at $x$. Choose $t=(R+r)/2$; then $R<t<r$. There is a point $y$ with $|y-a|=t$, for example $y=a+t$. Convergence Transfers Inward gives absolute convergence at $y$, since $t<r$. Thus $t\in D$, contradicting $t>R=\sup D$. The series must diverge at $x$, as claimed. $\square$

The strict inequalities are essential. The theorem makes no assertion when $|x-a|=R$. The set $D$ might contain $R$, or it might not; a series can also converge conditionally at a boundary point without that distance belonging to $D$. Determining the behavior at boundary points requires further analysis.

Calculating the Radius from the Coefficients

The definition describes the radius through convergence of a whole family of numerical series. The Root Test turns it into a calculation involving only the coefficients. Define the extended nonnegative number

L=\limsup_{n\to\infty}|c_n|^{1/n}.

The value $L$ measures the eventual exponential growth rate of the coefficients. The $n$th root of the absolute value of the $n$th term at distance $r>0$ is

\bigl(|c_n|r^n\bigr)^{1/n}=|c_n|^{1/n}r.

Consequently, the Root Test applies when $rL$ is strictly less than or strictly greater than $1$. This gives the Cauchy–Hadamard formula.

Theorem (Cauchy–Hadamard Formula): For a power series with coefficient growth rate $L=\limsup_{n\to\infty}|c_n|^{1/n}$, its radius of convergence is $$ R=\frac{1}{L}, $$ where $1/0=\infty$ and $1/\infty=0$.

Proof. First suppose $0<L<\infty$. Fix $r>0$. The Root Test applied to $\sum_{n=0}^{\infty}|c_n|r^n$ uses the limit superior

\limsup_{n\to\infty}\bigl(|c_n|r^n\bigr)^{1/n} = r\limsup_{n\to\infty}|c_n|^{1/n} = rL.

If $r<1/L$, then $rL<1$, so the Root Test gives absolute convergence at distance $r$. If $r>1/L$, then $rL>1$, so the Root Test gives divergence of the series of absolute values. In fact, the terms cannot tend to zero: their $n$th roots have limit superior greater than $1$, which would be impossible if the terms tended to zero. Thus the original series also diverges at any input at distance $r$. The Convergence Inside and Outside the Radius Theorem, or the definition of $R$, now gives $R=1/L$. No conclusion from the Root Test is needed when $r=1/L$.

If $L=0$, then for every finite $r>0$, the root limit superior $rL$ is $0<1$. The Root Test gives absolute convergence at every such distance. The center also converges, so $D$ is unbounded and $R=\infty=1/L$.

If $L=\infty$, then for every $r>0$, the root limit superior $rL$ is infinite and hence greater than $1$. The Root Test gives divergence at every positive distance. At distance zero the series converges, so $D=\{0\}$ and $R=0=1/L$. These cases complete the proof. $\square$

The formula is useful because it translates a question about convergence at many inputs into a question about the asymptotic size of one coefficient sequence. It also identifies a limitation of the Root Test: when $rL=1$, that test is inconclusive. This corresponds exactly to the boundary distance, where the radius theorem itself makes no claim.

Worked Examples: Finding Radii

Worked Example: Coefficients with a Fixed Exponential Growth Rate

Consider the power series centered at $1$

\sum_{n=0}^{\infty}3^n(x-1)^n.

Its coefficients are $c_n=3^n$, so for every $n\geq1$,

|c_n|^{1/n}=|(3^n)|^{1/n}=3.

Thus $L=3$, and the Cauchy–Hadamard formula gives $R=1/3$. For instance, at $x=1+1/6$, the distance from the center is $1/6<1/3$, and the terms form the geometric series $\sum_{n=0}^{\infty}(1/2)^n$, which converges absolutely. At $x=1+1/2$, the distance is $1/2>1/3$, and the terms are $(3/2)^n$, which do not tend to zero. This illustrates the inside and outside conclusions. The radius alone does not classify points exactly $1/3$ from the center.

Worked Example: Factorial Denominators Give Infinite Radius

Consider the series centered at $-2$

\sum_{n=0}^{\infty}\frac{(x+2)^n}{n!}.

For $n\geq1$, its coefficient satisfies $c_n=1/n!$. For $n\geq2$, at least $n/2$ of the factors in $n!=1\cdot2\cdots n$ are at least $n/2$. Hence

n!\geq\left(\frac{n}{2}\right)^{n/2}, \qquad |c_n|^{1/n}\leq\sqrt{\frac{2}{n}}.

The right-hand side tends to zero, so $L=0$. The Cauchy–Hadamard formula gives $R=\infty$, and the series converges absolutely for every real $x$. For example, at $x=5$, the distance from the center is $7$; the root-test quantity is still $7|c_n|^{1/n}$, which tends to zero. The Root Test therefore confirms absolute convergence there.

Worked Example: Factorial Coefficients Give Zero Radius

Now consider the series centered at $4$

\sum_{n=0}^{\infty}n!(x-4)^n.

For $n\geq2$, the same product estimate gives $n!\geq(n/2)^{n/2}$, and therefore

|n!|^{1/n}\geq\sqrt{\frac{n}{2}}.

This lower bound tends to infinity, so $L=\infty$ and $R=0$. At every $x\neq4$, put $r=|x-4|>0$. The $n$th-root quantity for the absolute terms is $r(n!)^{1/n}$, which tends to infinity. The Root Test shows divergence at every such input. At $x=4$, every term of positive index is zero and the remaining term is $0!=1$, so the series converges there. A zero radius does not mean the series fails everywhere; it means that no positive distance from the center is inside the radius.

Interpreting the Boundary

The radius divides the real line into three regions relative to the center: distances strictly less than $R$, the boundary distance $R$ when it is finite, and distances strictly greater than $R$. The first region consists entirely of points of absolute convergence. The last consists entirely of points of divergence. At the boundary, the coefficient growth rate alone may not settle the question, because the Root Test reaches the inconclusive value $1$.

A common mistake is to read “radius $R$” as saying the series converges whenever $|x-a|\leq R$. The theorem guarantees convergence only for the strict inequality $|x-a|<R$. For example, a radius can be zero while the series still converges at its center, as in the factorial-coefficient example. An infinite radius, by contrast, means that every real input is strictly inside the radius and the series converges absolutely everywhere.

Takeaway: A power series has a radius $R\in[0,\infty]$: it converges absolutely at every point with $|x-a|<R$ and diverges at every point with $|x-a|>R$. The coefficient growth formula is $R=1/\limsup |c_n|^{1/n}$, with the stated conventions for zero and infinity. Boundary behavior must be examined separately.

Check Your Understanding

Use the radius definition, the convergence theorem, and the Cauchy–Hadamard formula to answer the following questions.

Why does the series $\sum_{n=0}^{\infty}|c_n|r^n$ depend on the distance $r=|x-a|$, rather than on which side of the center $x$ lies?
If $R$ is finite and a power series converges at a point whose distance from the center is greater than $R$, which earlier result leads to a contradiction?
For coefficients $c_n=5^n$, compute $L$ and the radius of convergence.
What does $R=0$ guarantee about the series at its center, and what does it say about noncentral inputs?
Why does the Cauchy–Hadamard formula not by itself determine convergence at a point whose distance from the center equals a finite positive radius?

What You'll Learn