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Power Series · Tutorial 544 of 1000

The Cauchy-Hadamard Formula

Use the limsup of the coefficient roots to determine a power series’ radius, including for sparse coefficients and sequences whose roots do not converge.

Advanced 9 min read

What You'll Learn

  • Interpret the coefficient growth rate in the Cauchy-Hadamard Formula
  • Compute a limsup when coefficient roots oscillate or vanish at most indices
  • Determine the radius when the coefficient-root limsup is zero or infinite
  • Show why changing finitely many coefficients leaves the radius unchanged
  • Relate geometric rescaling of coefficients to a change in radius

From Coefficient Growth to Radius

The Interval of Convergence tutorial used the radius to identify the open interval where a power series converges absolutely, then treated the boundary points separately. The Cauchy-Hadamard Formula explains how to obtain that radius directly from the coefficients. Its key input is not usually the limit of the coefficient roots: it is their limit superior, which records the largest growth that persists arbitrarily far out in the sequence.

This distinction matters because coefficient roots may oscillate, or most coefficients may be zero. A limit may then fail to exist even though the power series still has a well-defined radius. The limsup handles these cases and gives a single coefficient-growth quantity to use in the formula.

The Formula and Its Meaning

For a power series \(\sum_{n=0}^{\infty}c_n(x-a)^n\), consider the sequence \(|c_n|^{1/n}\) for \(n\geq1\). The constant coefficient \(c_0\) is left out: one fixed term cannot determine the long-run growth of the coefficients. Recall the Cauchy-Hadamard Formula established earlier in this course.

Recall (Cauchy-Hadamard Formula): Let \(L=\limsup_{n\to\infty}|c_n|^{1/n}\). The radius of convergence is \(R=1/L\), with the conventions \(1/0=\infty\) and \(1/\infty=0\).

The limsup can be understood through tail suprema: for each index \(N\), take the supremum of all values \(|c_n|^{1/n}\) with \(n\geq N\), then let \(N\) increase. Values that occur only in an initial finite segment eventually disappear from these tails. By contrast, a large value that keeps recurring arbitrarily far out can affect the limsup, even if many other terms are smaller.

If the roots converge, their limit is also their limsup, so the computation may be straightforward. But if they do not converge, replacing the limsup by a nonexistent limit is not an option. The radius is determined by the tail behavior captured by the limsup, not by a typical coefficient or by a finite number of unusually large coefficients.

Worked Examples: Reading the Coefficient Roots

Worked Example: Coefficients Supported on Powers of Two

Define coefficients by \(c_{2^k}=3^{2^k}\) for integers \(k\geq0\), and set \(c_n=0\) at all other positive indices. At an index \(n=2^k\),

$$ |c_n|^{1/n}=(3^{2^k})^{1/2^k}=3. $$

At every other positive index, \(|c_n|^{1/n}=0\). The root sequence therefore does not converge: it equals \(3\) at powers of two and \(0\) at the other indices. There are arbitrarily large powers of two, so every tail contains a value of \(3\); no value in the sequence exceeds \(3\). Thus its limsup is \(3\), and the radius is \(R=1/3\).

The zero coefficients do not force the radius to be infinite. The nonzero coefficients recur at arbitrarily large indices, and their roots determine the limsup.

Worked Example: Oscillating Roots

Let \(c_n=4^n\) when \(n\) is even and \(c_n=2^n\) when \(n\) is odd, for \(n\geq1\). Then the roots are

$$ |c_n|^{1/n}= \begin{cases} 4,& n\text{ is even},\\ 2,& n\text{ is odd}. \end{cases} $$

These roots do not converge. Every tail contains even indices, where the value is \(4\), and every root is at most \(4\). Consequently, the limsup is \(4\). The Cauchy-Hadamard Formula gives \(R=1/4\). The smaller odd-index roots do not change this value because the larger roots occur arbitrarily far out.

Worked Example: Zero and Infinite Root Limsups

First take \(c_n=5^{-n^2}\) for \(n\geq1\). Then

$$ |c_n|^{1/n}=(5^{-n^2})^{1/n}=5^{-n}\longrightarrow0. $$

The limsup is \(0\), so the radius is infinite. The power series converges for every real \(x\), by the Convergence Inside and Outside the Radius theorem, since every finite distance from its center is less than an infinite radius.

Now take \(c_n=n^n\) for \(n\geq1\). In this case \(|c_n|^{1/n}=n\), which tends to infinity. The limsup is infinite, so the radius is \(0\). The series still converges at its center, as every power series does, but it diverges at every other real input by the same radius theorem.

Finite Coefficient Changes Do Not Affect the Radius

The tail-supremum interpretation yields a useful invariance principle. Changing a finite number of coefficients can change the power series by a polynomial, but it cannot change its radius. This makes it legitimate to simplify finitely many coefficients when calculating long-run growth, provided the coefficients beyond some index are unchanged.

Theorem (Finite Coefficient Changes Preserve the Radius): Suppose two power series have coefficient sequences \((c_n)\) and \((d_n)\), and there is an integer \(N\) such that \(c_n=d_n\) for every \(n\geq N\). Then the two power series have the same radius of convergence.

Proof. For every \(n\geq N\), the equality of coefficients gives \(|c_n|^{1/n}=|d_n|^{1/n}\). Thus the two sequences of roots agree on every sufficiently far tail. Their tail suprema are therefore equal once the tail begins at or beyond \(N\), so their limsups are equal. The Cauchy-Hadamard Formula assigns the same radius to equal limsups, including the cases where the common limsup is \(0\) or infinity. Hence the radii agree. \(\square\)

Worked Example: Simplifying Initial Coefficients

Suppose \(c_n=2^n\) for every \(n\geq4\), while \(c_0,c_1,c_2,c_3\) are arbitrary real numbers. For every \(n\geq4\),

$$ |c_n|^{1/n}=2. $$

Therefore \(L=2\) and the radius is \(1/2\). Replacing the first four coefficients by any other real numbers leaves all roots from index \(4\) onward unchanged. By the Finite Coefficient Changes Preserve the Radius theorem, the modified series also has radius \(1/2\). Its values and endpoint behavior can change, but its radius cannot.

Geometric Rescaling and the Radius

A second useful calculation concerns multiplying the \(n\)th coefficient by a geometric factor. This is the coefficient-level counterpart of changing the scale of the variable. The result lets us update a known radius without recomputing every coefficient root from scratch.

Theorem (Geometric Coefficient Rescaling): Let \(\lambda\neq0\), and define \(d_n=\lambda^n c_n\) for every \(n\geq0\). If the power series with coefficients \(c_n\) has radius \(R\), then the power series with coefficients \(d_n\) has radius \(R/|\lambda|\), using the usual extended-value conventions.

Proof. For each \(n\geq1\),

$$ |d_n|^{1/n}=|\lambda^n c_n|^{1/n}=|\lambda|\,|c_n|^{1/n}. $$

Multiplication by the fixed positive number \(|\lambda|\) multiplies every tail supremum by \(|\lambda|\), and therefore multiplies the limsup by \(|\lambda|\). If the original limsup is a finite positive number \(L\), the new one is \(|\lambda|L\), so the new radius is \(1/(|\lambda|L)=R/|\lambda|\). If \(L=0\), the new limsup is \(0\) and both radii are infinite. If \(L=\infty\), the new limsup is infinite and both radii are zero. These cases prove the claim in full. \(\square\)

Worked Example: Rescaling a Power Series

Start with coefficients \(c_n=3^n\). Their roots equal \(3\), so the radius is \(1/3\). Now define \(d_n=(1/2)^n c_n\). The rescaling theorem uses \(\lambda=1/2\), and gives the new radius

$$ \frac{1/3}{|1/2|}=\frac{2}{3}. $$

Directly, \(d_n=(3/2)^n\), and hence \(|d_n|^{1/n}=3/2\). The Cauchy-Hadamard Formula gives \(1/(3/2)=2/3\), confirming the result. This direct check also verifies that the transformed coefficients satisfy the claimed growth rate.

Using the Formula Carefully

The formula answers a precise question: what is the radius determined by the eventual coefficient growth? It does not classify convergence at the boundary. Once the radius is known, the Endpoint Classification of the Interval of Convergence theorem applies; any finite-radius endpoints still require separate tests.

A common mistake is to compute a few roots and treat their largest observed value as the limsup. A finite collection of values cannot determine tail behavior. Another is to assume that the roots must converge before the formula can be used. The sparse and oscillating examples show why neither approach is reliable. Instead, identify which root values occur arbitrarily far out and determine the supremum behavior of the tails.

The next tutorial develops the Ratio Test for power series, which can sometimes provide a more direct radius calculation when consecutive coefficient ratios have a limit. The Cauchy-Hadamard Formula remains useful even when those ratios fail to settle, because it is formulated in terms of limsup rather than a consecutive-ratio limit.

Takeaway: The radius is the reciprocal of the limsup of the coefficient roots. Tail behavior controls that limsup: finite initial changes do not affect it, while geometric rescaling multiplies it by the corresponding absolute scale factor.

Check Your Understanding

Use the coefficient-root limsup and the results in this tutorial to answer the following questions.

  1. If \(|c_n|^{1/n}\) alternates between \(1\) and \(6\), what is its limsup and the radius of the associated power series?
  2. Suppose \(c_n=0\) except at indices \(n=3^k\), where \(c_{3^k}=7^{3^k}\). What is the coefficient-root limsup?
  3. Why can changing finitely many coefficients not change the radius of convergence?
  4. If \(d_n=4^n c_n\) and the original radius is \(R\), what is the new radius?
  5. What do limsup values \(0\) and infinity imply about the radius?

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