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Power Series Structure · Tutorial 708 of 998

Deriving the Ratio-Test Radius Formula

Use the ratio test on consecutive power-series terms to derive the radius formula and identify its limitations.

Beginner 9 min read

What You'll Learn

  • Form the ratio of consecutive terms in a power series centered at a point.
  • State the condition on coefficient ratios required for the radius formula.
  • Derive the reciprocal relationship between the coefficient-ratio limit and the radius.
  • Interpret zero and infinite coefficient-ratio limits.
  • Recognize when a missing ratio limit leaves this method inconclusive.

Why Consecutive Terms Reveal a Radius

A power series is a sum whose terms depend on a variable through powers of its distance from a fixed center. For a series centered at \(c\), write

$$ \sum_{n=0}^{\infty} a_n(x-c)^n. $$

Here \(a_n\) is the coefficient of the \(n\)th power. The radius of convergence describes how far from \(c\) the series converges absolutely: points strictly inside the radius converge absolutely, while points strictly outside it diverge. The endpoints at exactly the radius require separate tests.

The ratio test gives a direct way to identify this dividing distance when consecutive coefficient magnitudes have a limit ratio. The key is to apply the test to the whole term, not just to its coefficient. The factor \((x-c)^n\) contributes a factor of \(|x-c|\) to the ratio, and that is what makes the result a radius.

Set Up the Ratio Test

Let \(u_n=a_n(x-c)^n\) be the \(n\)th term of the power series. If \(a_n\) and \(a_{n+1}\) are nonzero, the ratio of consecutive term magnitudes is

$$ \frac{|u_{n+1}|}{|u_n|} = \frac{|a_{n+1}(x-c)^{n+1}|}{|a_n(x-c)^n|} = \frac{|a_{n+1}|}{|a_n|}|x-c|, \qquad x\ne c. $$

The ratio test says that if this expression has a limit \(q<1\), the series converges absolutely; if its limit is greater than \(1\), or is infinite, the series diverges. If the limit is \(1\), the test does not decide convergence or divergence.

Condition for the coefficient-ratio formula: Assume the coefficients are nonzero for all sufficiently large \(n\), and that the limit \(L=\lim_{n\to\infty}\frac{|a_{n+1}|}{|a_n|}\) exists, with \(L\) allowed to be \(0\) or \(+\infty\). The formula for the radius is \(R=1/L\), using \(1/0=+\infty\) and \(1/(+\infty)=0\).

Requiring the coefficients to be eventually nonzero makes the consecutive coefficient ratios well-defined from some point onward. The absolute values matter because the ratio test applies to magnitudes; signs of coefficients do not alter this calculation.

Deriving the Radius Formula

First suppose \(0<L<\infty\), and fix a point \(x\ne c\). Since \(|x-c|\) is constant with respect to \(n\), the limit of the consecutive term ratio is

$$ \lim_{n\to\infty}\frac{|u_{n+1}|}{|u_n|} = \lim_{n\to\infty} \left(\frac{|a_{n+1}|}{|a_n|}|x-c|\right) = L|x-c|. $$

The ratio test therefore gives absolute convergence when \(L|x-c|<1\) and divergence when \(L|x-c|>1\). Since \(L>0\), dividing by \(L\) shows that these conditions are exactly

$$ |x-c|<\frac{1}{L} \qquad\text{and}\qquad |x-c|>\frac{1}{L}, $$

respectively. Thus the dividing distance from the center is \(1/L\). At the boundary, where \(|x-c|=1/L\), the term-ratio limit is \(1\), so the ratio test gives no endpoint conclusion.

Derivation: For a power series with an existing coefficient-ratio limit \(L\), the ratio test compares \(L|x-c|\) with \(1\). The transition occurs at distance \(1/L\), so the radius is \(R=1/L\).

This argument explains why the reciprocal appears: the coefficient ratio measures the eventual change in coefficient size from one index to the next, while the distance from the center contributes an additional factor. The product must be below \(1\) for the terms to shrink at the rate required by the ratio test.

At the center \(x=c\), every term with \(n\ge 1\) is zero, so the series reduces to its constant term \(a_0\). It converges there regardless of the coefficient-ratio limit. The ratio calculation above is needed for points away from the center.

Worked Examples

Worked Example: A Coefficient Ratio Approaching One

Find the radius of convergence of

$$ \sum_{n=0}^{\infty}\frac{1}{n+1}(x-2)^n. $$

Here \(a_n=\frac{1}{n+1}\), and every coefficient is nonzero. Compute the consecutive coefficient ratio:

$$ \frac{|a_{n+1}|}{|a_n|} = \frac{1/(n+2)}{1/(n+1)} = \frac{n+1}{n+2}. $$

Taking the limit gives \(L=1\). Therefore

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

Check the ratio at \(x=\frac{5}{2}\). The center is \(2\), so \(|x-2|=\frac12\), and the term-ratio limit is \(1\cdot\frac12=\frac12<1\), confirming absolute convergence there. At \(x=3\), the distance is \(1\), so the ratio limit is \(1\); the ratio test does not decide that boundary point. The radius calculation identifies the distance \(1\), not the endpoint behavior.

Worked Example: A Finite Radius Greater Than One

Find the radius of convergence of

$$ \sum_{n=0}^{\infty}\frac{n+1}{4^n}(x+1)^n. $$

The center is \(c=-1\), and \(a_n=\frac{n+1}{4^n}\). The coefficients are nonzero. Their consecutive ratio is

$$ \frac{|a_{n+1}|}{|a_n|} = \frac{(n+2)/4^{n+1}}{(n+1)/4^n} = \frac{n+2}{4(n+1)}. $$

Dividing numerator and denominator inside the fraction by \(n+1\) shows that

$$ L=\lim_{n\to\infty}\frac{n+2}{4(n+1)} = \frac14\lim_{n\to\infty}\frac{1+2/n}{1+1/n} = \frac14. $$

Hence the radius is

$$ R=\frac{1}{L}=\frac{1}{1/4}=4. $$

Check an interior point, \(x=1\). Its distance from \(-1\) is \(2\), so the term-ratio limit is \(\frac14\cdot2=\frac12<1\), confirming absolute convergence. At \(x=4\), the distance is \(5\), and the term-ratio limit is \(\frac14\cdot5=\frac54>1\), confirming divergence.

Worked Example: A Different Center and a Smaller Radius

Find the radius of convergence of

$$ \sum_{n=0}^{\infty}\frac{2n+1}{3^n}(x-1)^n. $$

The center is \(c=1\), with \(a_n=\frac{2n+1}{3^n}\). The consecutive coefficient ratio is

$$ \frac{|a_{n+1}|}{|a_n|} = \frac{(2n+3)/3^{n+1}}{(2n+1)/3^n} = \frac{2n+3}{3(2n+1)}. $$

Dividing the numerator and denominator by \(n\) gives

$$ L=\lim_{n\to\infty}\frac{2n+3}{3(2n+1)} = \frac{2}{3\cdot2} = \frac13. $$

Therefore the radius is

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

As a check, take \(x=2\), which is distance \(1\) from the center. The term-ratio limit is \(\frac13\cdot1=\frac13<1\), so the series converges absolutely there. At \(x=5\), the distance is \(4\), giving ratio limit \(\frac13\cdot4=\frac43>1\), so the series diverges there.

Zero and Infinite Limits

The same comparison describes the extreme cases. If \(L=0\), then for every fixed finite distance \(|x-c|\), the term-ratio limit is \(0\cdot|x-c|=0<1\). The series converges absolutely at every real \(x\), and the radius is infinite. If \(L=+\infty\), then for every \(x\ne c\), the term-ratio limit is infinite, so the series diverges at every point away from the center. It still converges at \(x=c\), and its radius is zero.

These conclusions require the stated coefficient-ratio limit. A polynomial, whose coefficients are eventually zero, is a separate straightforward case: all but finitely many terms vanish for every \(x\), so it converges everywhere. Its coefficient ratios are not eventually defined, and the ratio formula is not the reason for its infinite radius.

When the Coefficient-Ratio Limit Does Not Exist

If \(\frac{|a_{n+1}|}{|a_n|}\) has no limit, the formula \(R=1/L\) cannot be applied because there is no value \(L\) to substitute. For any fixed \(x\ne c\), multiplying these ratios by the fixed positive number \(|x-c|\) does not create a limit if the coefficient ratios have none. Thus the ordinary ratio test, in this direct form, does not determine convergence at such a point.

This is a limitation of this method, not proof that the power series has no radius of convergence. A series can have a radius even when consecutive coefficient ratios fail to settle to one limit. Other reasoning may still determine its convergence behavior. Also, if the coefficient ratios exist but the resulting term-ratio limit equals \(1\) at a particular point, that is a boundary case where the ratio test is inconclusive, not a failure of the radius formula itself.

Worked Example: Correcting a Reciprocal Error

For the series in the second worked example, the coefficient-ratio limit is \(L=\frac14\). A common incorrect calculation is to report \(R=\frac14\), treating the coefficient ratio itself as the radius.

Check the actual term ratio instead. For \(a_n=\frac{n+1}{4^n}\) and center \(c=-1\), it is

$$ \lim_{n\to\infty}\frac{|a_{n+1}(x+1)^{n+1}|}{|a_n(x+1)^n|} = \frac14|x+1|. $$

For convergence, this quantity must be less than \(1\):

$$ \frac14|x+1|<1 \quad\Longleftrightarrow\quad |x+1|<4. $$

The distance threshold is \(4\), so the correct radius is \(R=4=1/L\), not \(1/4\). Substituting the corrected value into the term-ratio condition gives \(\frac14 R=1\), exactly the boundary transition required by the ratio test.

Why the Formula Is Useful

The formula compresses a convergence calculation into two steps: find the limiting ratio of consecutive coefficient magnitudes, then take its reciprocal. The center determines where the interval of convergence is centered, while the coefficient ratios determine its radius. This distinction is useful because changing the center changes the expression \((x-c)^n\), but the radius calculation depends on the coefficient growth and the distance factor.

Keep the conditions visible when using the shortcut. The coefficients must be eventually nonzero for consecutive ratios to be defined, and their absolute-value ratio must have a limit. A finite positive limit gives a finite positive radius. A zero limit gives infinite radius; an infinite limit gives radius zero. If the limit is absent, or if the term-ratio limit is \(1\) at a point, this ratio-test argument does not settle the question. In particular, the formula gives the radius, not a decision about convergence at either boundary point.

Check Your Understanding

Use the ratio-test reasoning developed above to answer these questions.

  1. For a power series centered at \(c\), what factor involving \(x\) appears in the ratio of consecutive term magnitudes?
  2. If the coefficient-ratio limit is \(L=2\), what radius does the formula give?
  3. Why does a term-ratio limit of \(1\) at a boundary point not determine whether that point is included?
  4. What can be concluded from the coefficient-ratio formula if the limit is \(0\)?
  5. If the coefficient-ratio limit does not exist, does that prove that the power series has no radius of convergence? Explain.

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