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Power Series Structure · Tutorial 710 of 998

Computing Radius from Consecutive Coefficients

Use consecutive coefficient ratios to find a power series’ radius, and learn when the ratio test applies and what it leaves undecided.

Beginner 10 min read

What You'll Learn

  • State the conditions under which a coefficient-ratio limit determines the radius.
  • Relate the coefficient-ratio limit to the ratio test at points away from the center.
  • Compute radii for reciprocal-factorial and reciprocal-power coefficients.
  • Simplify ratios involving products of factorial factors.
  • Identify why the coefficient ratio does not settle convergence at boundary points.
  • Avoid applying the term-ratio formula at the center of a power series.

Why Consecutive Coefficients Reveal the Radius

A power series has terms whose sizes depend on both its coefficients and the distance from its center. The coefficients may be complicated, but comparing one coefficient with the next can make their long-term growth much easier to recognize.

When the absolute ratio of consecutive coefficients approaches a limit, the ratio test turns that limit into a threshold for the distance from the center. This tutorial develops that connection and applies it to reciprocal factorials, reciprocal powers, and products of factorial factors.

The Coefficient-Ratio Formula

Consider a power series centered at \(c\):

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

Assume that its coefficients are nonzero for all sufficiently large \(n\), so the consecutive coefficient ratios are defined eventually. Suppose the limit

$$ L=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| $$

exists, allowing \(L=0\) or \(L=+\infty\). When \(0<L<\infty\), the radius of convergence is the reciprocal of this limit. The conventions for the limiting cases are included below.

Theorem: If the coefficients are eventually nonzero and \(L=\lim_{n\to\infty}|a_{n+1}/a_n|\) exists, then the radius of convergence \(R\) is \(1/L\) when \(0<L<\infty\). If \(L=0\), then \(R=\infty\). If \(L=\infty\), then \(R=0\). This conclusion specifies convergence for points strictly inside or outside the radius; it does not, by itself, decide convergence at boundary points.

The ratio in this theorem is a ratio of coefficients. For a fixed \(x\ne c\), the ratio of the absolute values of consecutive terms is

$$ \left|\frac{a_{n+1}(x-c)^{n+1}}{a_n(x-c)^n}\right| = \left|\frac{a_{n+1}}{a_n}\right||x-c|. $$

The condition \(x\ne c\) matters: it allows cancellation of the powers of \(x-c\), and it ensures those powers are nonzero. At \(x=c\), handle the series directly instead. Every term with \(n\ge 1\) then vanishes, so the series converges at its center.

Why the Formula Works

For a fixed \(x\ne c\), call the absolute value of the \(n\)th term \(u_n=|a_n(x-c)^n|\). Since the coefficients are eventually nonzero, these terms are eventually positive, and

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

The ratio test says that a series of positive terms converges if the limit of its successive-term ratios is less than \(1\), and diverges if that limit is greater than \(1\). When the limit equals \(1\), the test gives no conclusion.

1
Compare the distance with the limiting ratio.
For \(0<L<\infty\), the term-ratio limit is \(L|x-c|\). It is less than \(1\) exactly when \(|x-c|<1/L\).
2
Apply the ratio test inside the threshold.
If \(|x-c|<1/L\), then \(L|x-c|<1\), so the power series converges absolutely.
3
Apply the ratio test outside the threshold.
If \(|x-c|>1/L\), then \(L|x-c|>1\), so the terms do not tend to zero and the series diverges.

Thus the threshold distance is \(1/L\), which gives the radius. At equality, \(|x-c|=1/L\), the term-ratio limit is \(1\), so this argument does not settle convergence there.

If \(L=0\), then for every fixed \(x\ne c\), the term-ratio limit is \(0\), so the series converges absolutely; it also converges at \(c\). Hence \(R=\infty\). If \(L=\infty\), then for each \(x\ne c\), the term ratios eventually exceed a number greater than \(1\). The terms cannot tend to zero, so the series diverges at every such point. It converges at \(c\), giving \(R=0\).

Worked Examples

Worked Example: Reciprocal Factorial Coefficients

Find the radius of convergence of

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

Here \(c=2\) and \(a_n=1/n!\). The factorial ratio is

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

Therefore,

$$ L=\lim_{n\to\infty}\frac{1}{n+1}=0, \qquad R=\infty. $$

To check using the ratio test, fix \(x\ne2\). The ratio of the absolute values of consecutive terms is \(|x-2|/(n+1)\), which tends to \(0\). The series therefore converges absolutely for every \(x\ne2\). At \(x=2\), all terms with \(n\ge1\) are zero and the \(n=0\) term is \(1\), so the series converges there as well. The result is convergence for every real \(x\), as required by \(R=\infty\).

Worked Example: Reciprocal-Power Coefficients

Find the radius of convergence of

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

The center is \(c=-1\), and \(a_n=1/(n+1)^2\). Consecutive coefficients satisfy

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

Since \((n+1)/(n+2)\) tends to \(1\),

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

For a direct check, take any \(x\ne-1\). The ratio of consecutive absolute term values is

$$ \left(\frac{n+1}{n+2}\right)^2|x+1| \longrightarrow |x+1|. $$

If \(|x+1|<1\), the limit is less than \(1\), so the series converges absolutely. If \(|x+1|>1\), the limit is greater than \(1\), so it diverges. At \(x=-1\), only the constant term remains, so the series converges there. These checks confirm radius \(1\); at \(|x+1|=1\), this ratio calculation alone does not decide convergence.

Worked Example: A Product of Factorial Factors

Find the radius of convergence of

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

Here \(c=0\) and \(a_n=(n!)^2/(2n)!\). Compute the next coefficient and divide:

$$ \left|\frac{a_{n+1}}{a_n}\right| = \frac{((n+1)!)^2}{(2n+2)!}\frac{(2n)!}{(n!)^2} = \frac{(n+1)^2}{(2n+2)(2n+1)}. $$

Using \(2n+2=2(n+1)\), simplify before taking the limit:

$$ \frac{(n+1)^2}{(2n+2)(2n+1)} = \frac{(n+1)^2}{2(n+1)(2n+1)} = \frac{n+1}{2(2n+1)} \longrightarrow \frac14. $$

Thus \(L=1/4\), and the radius is

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

For \(x\ne0\), the consecutive absolute term ratio tends to \(|x|/4\). It is less than \(1\) when \(|x|<4\), so the series converges absolutely there, and greater than \(1\) when \(|x|>4\), so the series diverges there. At \(x=0\), only the \(n=0\) term remains and equals \(1\). This verifies the radius \(4\); the ratio limit is \(1\) at \(x=4\) and \(x=-4\), so it does not determine convergence at either point.

Common Errors and Their Corrections

A frequent mistake is to reverse the coefficient-ratio limit. For example, if \(L=1/4\), the radius is not \(1/4\). The term-ratio limit at a point \(x\ne c\) is \((1/4)|x-c|\), and the ratio test requires this to be less than \(1\):

$$ \frac14|x-c|<1 \quad\Longleftrightarrow\quad |x-c|<4. $$

The threshold distance, and therefore the radius, is \(4=1/L\).

Another error is to substitute the center into the quotient of consecutive terms. In the reciprocal-factorial example, someone might write that the term ratio is \(|x-2|/(n+1)\) “for every \(x\),” then substitute \(x=2\) into the original quotient. At \(x=2\), both consecutive terms with positive indices are zero, so that quotient is \(0/0\), which is undefined. The correction is to state the term-ratio formula only for \(x\ne2\), and then check \(x=2\) directly: all positive-power terms vanish, leaving the finite constant term.

Takeaway: The coefficient-ratio limit determines the radius when it exists under the stated nonzero-coefficient condition. Use the ratio test away from the center, check the center directly, and do not treat a ratio limit of \(1\) at the boundary as a convergence verdict.

A Reliable Calculation Process

For a power series with coefficients that are products or quotients, organize the calculation before taking limits:

  1. Identify the center \(c\) and write down the coefficient \(a_n\) separately from \((x-c)^n\).
  2. Form \(|a_{n+1}/a_n|\), keeping the index \(n\) visible in every factorial and power.
  3. Simplify the expression exactly before taking its limit.
  4. Set \(R=1/L\) for a positive finite limit, and use the stated conventions when the limit is zero or infinite.
  5. Interpret the result using the ratio test only for \(x\ne c\); check \(x=c\) directly.

This method is useful because factorials and powers often look large while their consecutive ratios simplify to a short expression. It also separates two different questions: the coefficient calculation determines the radius, while convergence at points exactly on the boundary requires a separate analysis.

Check Your Understanding

Use the coefficient-ratio method, and remember to distinguish the center from points where the term ratio is defined.

  1. For \(\sum_{n=0}^{\infty}(x-3)^n/(n+2)!\), calculate the limiting absolute coefficient ratio and the radius.
  2. For \(\sum_{n=0}^{\infty}(x+2)^n/(n+1)^4\), find the radius and state what the ratio test says at points strictly inside and outside it.
  3. For \(\sum_{n=0}^{\infty}((n!)^2/(2n)!) (x-1)^n\), simplify the coefficient ratio and determine the radius.
  4. Why is the consecutive-term quotient not defined at the center, even though the series converges there?
  5. If the coefficient-ratio limit is \(1\), what radius follows, and what does that calculation say about boundary convergence?

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