From a Ratio Limit to a Radius
A power series has a center and coefficients, and its terms change as the input moves away from that center. The ratio test turns that change into a condition on the distance from the center. When the limit of consecutive coefficient ratios is a positive finite number, the test gives a precise radius: inside it, the series converges absolutely; outside it, the series diverges.
The key is to apply the ratio test to the whole term, including the power of the distance from the center. The resulting inequality is then solved for that distance. At the boundary, the ratio test itself is inconclusive, so a radius calculation does not by itself determine whether either endpoint is included.
Power Series and Radius of Convergence
A power series centered at \(a\) is an infinite series of the form
where \(a\) is the center, \(c_n\) is the coefficient of the \(n\)th power, and \(x\) is the input. Its general term is \(u_n(x)=c_n(x-a)^n\). For a fixed \(x\), the question is whether the numerical series made from these terms converges.
To find \(R\) with the ratio test, suppose the coefficients are eventually nonzero and the following limit exists and is positive and finite:
For \(x\ne a\), the ratio of consecutive term magnitudes is
Since \(|x-a|\) is fixed as \(n\) varies, the limit of this term ratio is \(\lambda|x-a|\). The ratio test says that a series converges absolutely when this limit is less than \(1\), and diverges when it is greater than \(1\). Thus the condition for absolute convergence is \(\lambda|x-a|<1\), or \(|x-a|<1/\lambda\). This makes the radius \(1/\lambda\).
Why the Formula Works
The radius formula follows directly from the ratio test and the meaning of distance from the center. In particular, the same coefficient limit applies at every fixed input; only the factor \(|x-a|\) changes. The following argument shows how the inside and outside regions arise.
For fixed \(x\ne a\), the limit of \(\left|u_{n+1}(x)/u_n(x)\right|\) is \(\lambda|x-a|\).
If \(\lambda|x-a|<1\), the ratio test gives absolute convergence. Dividing by \(\lambda>0\) gives \(|x-a|<1/\lambda\).
If \(\lambda|x-a|>1\), the ratio test gives divergence. Dividing by \(\lambda>0\) gives \(|x-a|>1/\lambda\).
Convergence is established for distances less than \(1/\lambda\), and divergence for distances greater than \(1/\lambda\). Therefore the radius is \(R=1/\lambda\); the equality case is not decided by the ratio test.
At the center \(x=a\), every term with \(n\ge 1\) is zero, so the series converges there. The ratio calculation above uses \(x\ne a\) because dividing consecutive terms can involve powers of \(x-a\). The conclusion about the radius includes the center as part of the convergence region.
Worked Examples
Worked Example: A Denominator That Grows Linearly
Find the radius of convergence of
The center is \(a=2\), and the coefficient is \(c_n=1/[3^n(n+1)]\). Calculate the ratio of consecutive coefficient magnitudes:
Divide the numerator and denominator of the fraction involving \(n\) by \(n\), or use the leading terms, to get
For a fixed \(x\ne2\), the ratio of consecutive term magnitudes has limit
The ratio test gives absolute convergence when \(|x-2|/3<1\). Multiplying by \(3\) gives \(|x-2|<3\). Therefore the radius is \(R=3\). Check the boundary threshold by substituting \(R=3\): the limiting ratio there is \(3/3=1\), exactly the equality case for which the ratio test is inconclusive. The radius calculation does not decide whether the series converges at either boundary input.
Worked Example: Binomial Coefficients and a Shifted Center
Find the radius of convergence of
The center is \(a=-1\), and \(c_n=(2n)!/(n!)^2\). Form the consecutive coefficient ratio:
As \(n\) grows, the numerator \(2(2n+1)\) and denominator \(n+1\) have leading terms \(4n\) and \(n\), so
The consecutive term-magnitude ratio therefore has limit \(4|x+1|\). The ratio test gives convergence when \(4|x+1|<1\). Dividing by \(4\) yields \(|x+1|<1/4\), so \(R=1/4\). Check: at a distance of \(1/4\) from the center, the limiting ratio is \(4(1/4)=1\), as required for the boundary threshold. This confirms the radius but says nothing about convergence at those boundary inputs.
Worked Example: A Binomial-Coefficient Ratio with Several Factors
Find the radius of convergence of
The center is \(a=4\). It is convenient to include the factor \(8^{-n}\) in the coefficient, so let \(c_n=(3n)!/[n!(2n)!8^n]\). Then
The numerator factors have leading terms \(3n,3n,3n\), while the denominator factors have leading terms \(n,2n,2n\). Hence
The general term is \(c_n(x-4)^n\), so the term-ratio limit is \((27/32)|x-4|\). The ratio test gives convergence if
Thus \(R=32/27\). Check the threshold: at distance \(32/27\), the ratio limit is \((27/32)(32/27)=1\). The result is consistent with the ratio-test equality case, which is inconclusive.
A Common Mistake: Treating Equality as Convergence
A frequent error is to calculate the ratio limit correctly, set it equal to \(1\), and then conclude that the series converges at the boundary. But the ratio test has three outcomes: a limit less than \(1\) guarantees absolute convergence; a limit greater than \(1\) guarantees divergence; a limit equal to \(1\) gives no conclusion.
Worked Example: Correcting a Boundary Conclusion
For the first series, the term-ratio limit is \(|x-2|/3\). At \(x=5\), the distance from the center is \(|5-2|=3\), so
Incorrect conclusion: “The ratio limit is \(1\), so the series converges at \(x=5\).” This does not follow from the ratio test. The correct conclusion is that this test is inconclusive at \(x=5\). Determining convergence at a boundary input requires a separate examination of the series at that input. The radius itself remains \(3\), regardless of whether the boundary series converges.
Using the Radius Result Carefully
Once a positive finite coefficient-ratio limit has been found, the radius gives a concise description of what the ratio test establishes:
| Distance from the center | Ratio-test conclusion |
|---|---|
| \(|x-a|<R\) | The power series converges absolutely. |
| \(|x-a|>R\) | The power series diverges. |
| \(|x-a|=R\) | The ratio test is inconclusive; test the boundary separately. |
The distance is \(|x-a|\), not \(x-a\) alone: inputs to the left and right of the center are measured by their absolute distance. The radius is also not the same thing as an interval with specified endpoints. The radius identifies the boundary distance; further testing determines what happens at those boundary inputs.
For a routine calculation, keep the steps in this order: identify \(c_n\) and \(a\), find the limit of \(|c_{n+1}/c_n|\), multiply that limit by \(|x-a|\), and solve the ratio-test inequality. Finally, state the radius and explicitly note that equality is unresolved by the ratio test. Keeping these steps separate helps prevent algebra errors and overclaiming.
Check Your Understanding
For each question, apply the ratio test to the general term and distinguish the radius from any boundary conclusion.
- Find the radius of convergence of \(\displaystyle\sum_{n=0}^{\infty}(n+1)(x-3)^n/6^n\). What does the ratio test say when \(|x-3|\) equals the radius?
- For \(\displaystyle\sum_{n=0}^{\infty}(x+2)^n/[5^n(n+1)]\), find the limit of the consecutive coefficient ratio and the radius.
- A power series centered at \(a=1\) has \(\displaystyle\lim_{n\to\infty}|c_{n+1}/c_n|=2\). What is its radius, and for which distances does the ratio test guarantee absolute convergence?
- For \(\displaystyle\sum_{n=0}^{\infty}\frac{(2n)!}{(n!)^2}(x-1)^n\), compute the limit of \(|c_{n+1}/c_n|\) and find the radius.
- Explain why a ratio-test limit of \(1\) at \(|x-a|=R\) does not establish either convergence or divergence at that boundary input.
