The Root Test and Cauchy-Hadamard Formula

Why Coefficient Growth Determines the Radius

A power series has the form $\sum_{n=0}^{\infty}a_n(x-c)^n$, where $c$ is its center and $a_n$ are its coefficients. Its terms combine two kinds of growth: the coefficients $a_n$, and the powers of the distance $|x-c|$. The root test separates these effects. Taking nth roots turns the power $|x-c|^n$ into the fixed factor $|x-c|$, which allows us to read the convergence radius from the growth of the coefficients alone.

The quantity needed to measure coefficient growth is a limit superior. It is useful even when the sequence $|a_n|^{1/n}$ does not converge. This makes the resulting formula applicable to every power series, including ones whose coefficients fluctuate.

Definition: For a power series $\sum_{n=0}^{\infty}a_n(x-c)^n$, define $$ L=\limsup_{n\to\infty}|a_n|^{1/n}, $$ where the limit superior is allowed to be $+\infty$. The $n=0$ coefficient does not affect this quantity. The Cauchy–Hadamard formula defines the radius of convergence by $$ R=\frac{1}{L}, $$ with the conventions $1/0=+\infty$ and $1/(+\infty)=0$.

For this formula, the limit superior is taken over indices $n\ge 1$. If a coefficient is zero, its nth root is zero. The radius describes where the power series converges absolutely: it converges absolutely when $|x-c|<R$ and diverges when $|x-c|>R$. The cases $R=0$ and $R=+\infty$ have their natural meanings: only the center is guaranteed to converge in the first case, while every finite real $x$ is inside the radius in the second.

Deriving the Formula from the Root Test

Fix a real number $x\ne c$. The nth term of the series is $a_n(x-c)^n$. Its nth root in absolute value is

\left|a_n(x-c)^n\right|^{1/n} = |a_n|^{1/n}|x-c|.

Since $|x-c|$ is a fixed positive number, the limit superior of the right-hand side is $L|x-c|$, with the product interpreted in the extended nonnegative reals. The root test says that a series converges absolutely if the limit superior of the nth roots of its terms is less than $1$, and diverges if that limit superior is greater than $1$. If it equals $1$, the root test gives no conclusion.

The following derivation makes explicit how this criterion produces the radius.

Take nth roots of the terms.
For $x\ne c$, the root-test quantity is $\limsup_{n\to\infty}|a_n|^{1/n}|x-c|=L|x-c|$, because $|x-c|$ does not depend on $n$.

Apply the root-test thresholds.
If $L|x-c|<1$, the series converges absolutely. If $L|x-c|>1$, it diverges.

Translate the inequalities into distance from the center.
When $0<L<+\infty$, the inequalities become $|x-c|<1/L$ for absolute convergence and $|x-c|>1/L$ for divergence. Thus the radius is $R=1/L$.

This argument also covers the extreme coefficient-growth cases, with a small care at the center. If $L=0$, then for every finite $x\ne c$, $L|x-c|=0<1$, so the series converges absolutely at every such point; it also converges at $x=c$. Thus $R=+\infty$. If $L=+\infty$, then for every $x\ne c$, $L|x-c|=+\infty>1$, so the series diverges there. At the center, however, all terms with $n\ge1$ are zero, so the series converges there and $R=0$.

Theorem: For $\sum_{n=0}^{\infty}a_n(x-c)^n$, let $L=\limsup_{n\to\infty}|a_n|^{1/n}$. Its radius of convergence is $R=1/L$, with $1/0=+\infty$ and $1/(+\infty)=0$. When $0<R<+\infty$, the formula gives absolute convergence for $|x-c|<R$ and divergence for $|x-c|>R$; it does not decide convergence when $|x-c|=R$.

Worked Examples

Worked Example: A Finite Positive Radius

Find the radius of convergence of $\sum_{n=0}^{\infty}3^n(x-2)^n$.

Here $a_n=3^n$. For every $n\ge1$,

|a_n|^{1/n}=|3^n|^{1/n}=3.

Therefore $L=\limsup_{n\to\infty}3=3$, and the formula gives

R=\frac{1}{L}=\frac{1}{3}.

To check the result directly with the root test, for $x\ne2$ the nth root of the absolute value of the nth term is

\left|3^n(x-2)^n\right|^{1/n}=3|x-2|.

This is less than $1$ when $|x-2|<1/3$ and greater than $1$ when $|x-2|>1/3$, matching the computed radius. At $x=2$, all terms except the $n=0$ term are zero, so the series converges there as well. The root test gives no conclusion at the two points where $|x-2|=1/3$.

Worked Example: An Infinite Radius

Find the radius of convergence of $\sum_{n=0}^{\infty}2^{-n^2}(x+1)^n$.

The coefficients are $a_n=2^{-n^2}$. For $n\ge1$,

|a_n|^{1/n} = \left(2^{-n^2}\right)^{1/n} = 2^{-n}.

As $n\to\infty$, $2^{-n}\to0$, so $L=0$. The reciprocal convention therefore gives

R=\frac{1}{0}=+\infty.

For a direct check at any fixed real $x$, the nth root of the absolute value of the nth term is $2^{-n}|x+1|$, which tends to $0$. Since $0<1$, the root test gives absolute convergence. This includes $x=-1$, where the terms with $n\ge1$ are zero. Thus the infinite radius is consistent with the root-test conclusion at every finite $x$.

Worked Example: A Zero Radius

Find the radius of convergence of $\sum_{n=0}^{\infty}2^{n^2}x^n$.

In this case $a_n=2^{n^2}$. For $n\ge1$,

|a_n|^{1/n} = \left(2^{n^2}\right)^{1/n} = 2^n.

Since $2^n\to+\infty$, the limit superior is $L=+\infty$, and

R=\frac{1}{+\infty}=0.

For any $x\ne0$, the nth root of the absolute value of the nth term is $2^n|x|$, which tends to $+\infty$. It is eventually greater than $1$, so the root test gives divergence. At $x=0$, every term with $n\ge1$ is zero, leaving only the finite $n=0$ term, so the series converges. This verifies that convergence occurs at the center and nowhere else.

What the Boundary Case Means

For a finite positive radius $R$, the boundary consists of points satisfying $|x-c|=R$. At such a point, $L=1/R$, so the root-test quantity is

L|x-c|=\frac{1}{R}R=1.

The root test is inconclusive when its limit superior is $1$: it neither proves convergence nor proves divergence. Each boundary point must be assessed by a suitable test or argument if its convergence is needed. The radius formula alone does not include that information.

This boundary statement has important qualifications. If $R=0$, then $L=+\infty$, and the only point with $|x-c|=R$ is the center $x=c$. At that point the terms with $n\ge1$ are zero, so the series converges; it is not a case where the root-test quantity equals $1$. If $R=+\infty$, there are no finite boundary points to assess.

A Common Pitfall: Using the Formula at the Center

One must not substitute $x=c$ into $L|x-c|$ when $L=+\infty$ and treat the result as an ordinary product. It has the indeterminate form $+\infty\cdot0$. For example, the series in the zero-radius example has $L=+\infty$ and $c=0$. The incorrect calculation $L|c-c|=(+\infty)\cdot0=1$ would falsely suggest that the root test is inconclusive at the center.

The correction is to compute the terms at the center directly. For every $n\ge1$,

a_n(c-c)^n=a_n\cdot0^n=0.

Thus the series at $x=c$ consists only of its finite constant term and converges. The expression $L|x-c|$ was used in the derivation for $x\ne c$; at the center, direct substitution into the original series is the appropriate check.

How to Use the Formula Reliably

When applying Cauchy–Hadamard, keep the coefficient calculation separate from the point $x$. First compute the limit superior of $|a_n|^{1/n}$; then take its reciprocal, using the stated conventions for zero and infinity. Only after finding $R$ should you classify points by their distance from the center. This separation helps prevent index mistakes and avoids confusing coefficient growth with the behavior of a particular term.

The formula matters because it supplies a radius even when the sequence of coefficient roots has no ordinary limit. The limit superior records the eventual upper growth rate, and the root test needs precisely that information. It gives a complete answer about points strictly inside and strictly outside the radius, while leaving boundary behavior to separate analysis when the radius is finite and positive.

Check Your Understanding

Use the definition of $L$ and the Cauchy–Hadamard formula to answer these questions.

If $\limsup_{n\to\infty}|a_n|^{1/n}=4$, what is the radius of convergence?
What does $L=0$ imply about convergence at finite real values of $x$?
For $0<R<+\infty$, what is the root-test quantity at a point satisfying $|x-c|=R$, and what conclusion does the root test give?
If $L=+\infty$, why should convergence at $x=c$ be checked directly rather than by treating $L|x-c|$ as a usual product?
For the series $\sum_{n=0}^{\infty}5^n(x-1)^n$, compute $L$ and $R$.

The Root Test and Cauchy-Hadamard Formula

What You'll Learn

Why Coefficient Growth Determines the Radius

Deriving the Formula from the Root Test