Radius Is Not the Same as Interval
A power series has a center and a radius of convergence. The radius describes a distance from the center: at points strictly closer than that distance, the series converges absolutely; at points farther away, it diverges. The radius alone does not tell us whether the two points exactly that distance from the center are included.
For a real power series centered at \(c\), when its radius \(R\) is finite and positive, the two boundary points are \(c-R\) and \(c+R\). These are the endpoints that must be tested separately. The final interval of convergence depends on the results of those tests.
The radius is a nonnegative number; the interval is a set of real numbers. For example, saying that a series has radius \(3\) does not by itself say whether its interval is open, closed, or half-open. The endpoint tests supply that missing information.
Finding the Boundary Points
The interior points are those whose distance from the center is less than \(R\). In symbols, the condition is \(|x-c|<R\). Solving this double inequality gives the open interval from \(c-R\) to \(c+R\).
This calculation locates the boundary; it does not decide whether either boundary point is part of the convergence set. At the boundaries, the distance condition becomes equality:
The result for power series is that they converge absolutely inside their radius and diverge outside it. At equality, the interior-versus-exterior result does not apply. Substitute each boundary value into the series, simplify the resulting terms, and use an appropriate convergence test on each resulting series.
Worked Example: Neither Endpoint Is Included
Worked Example: A Geometric Series Centered at 2
Consider the power series \(\sum_{n=0}^{\infty}\frac{(x-2)^n}{3^n}\). Its terms form a geometric series with ratio \(\frac{x-2}{3}\). A geometric series converges when the absolute value of its ratio is less than \(1\), so its interior is determined by:
Thus the center is \(c=2\), the radius is \(R=3\), and the boundary points are \(2-3=-1\) and \(2+3=5\).
At the left endpoint, substitute \(x=-1\):
The terms \((-1)^n\) do not approach \(0\), so the series diverges by the term test. At the right endpoint, substitute \(x=5\):
Here too the terms do not approach \(0\), so the series diverges. Neither endpoint is included, and the interval of convergence is \((-1,5)\). The checks agree with the geometric-series condition: at each endpoint the ratio has absolute value \(1\), not less than \(1\).
Endpoint Tests Can Give Different Results
The two endpoint substitutions do not have to lead to series with the same behavior. The signs and magnitudes of the resulting terms can differ. One endpoint might yield an alternating series that converges, while the other yields a positive harmonic series that diverges.
Worked Example: The Left Endpoint Is Included
Consider \(\sum_{n=1}^{\infty}\frac{(x+2)^n}{n}\), with center \(c=-2\) and radius \(R=1\). Its boundary points are \(-2-1=-3\) and \(-2+1=-1\), and its interior is \((-3,-1)\).
At the left endpoint \(x=-3\), the power \(x+2\) becomes \(-1\):
This alternating series converges: its positive term magnitudes \(1/n\) decrease, since \(1/(n+1)<1/n\) for every \(n\geq1\), and \(\lim_{n\to\infty}1/n=0\). It does not converge absolutely because the series of absolute values is \(\sum_{n=1}^{\infty}1/n\), the divergent harmonic series. Thus the left endpoint is included, with conditional convergence.
At the right endpoint \(x=-1\), the power \(x+2\) becomes \(1\):
This is the harmonic series, so it diverges. Therefore the right endpoint is excluded, and the interval of convergence is \([-3,-1)\). The left bracket records convergence at \(-3\); the right parenthesis records divergence at \(-1\).
Worked Example: Both Endpoints Are Included
Worked Example: Convergence at Both Boundary Points
Consider \(\sum_{n=1}^{\infty}\frac{(x-1)^n}{n^2}\), whose center is \(c=1\) and radius is \(R=1\). The boundary points are \(1-1=0\) and \(1+1=2\), so the interior is \((0,2)\).
At \(x=0\), substitute \(x-1=-1\):
The series converges absolutely because its absolute-value series is \(\sum_{n=1}^{\infty}1/n^2\), a convergent \(p\)-series with \(p=2>1\).
At \(x=2\), substitute \(x-1=1\):
This is again a \(p\)-series with \(p=2>1\), so it converges. Both endpoints are included, and the interval of convergence is \([0,2]\). In this example both endpoint series converge absolutely.
How to Write the Interval Correctly
Use a square bracket exactly where the endpoint series converges, and a parenthesis where it diverges. The center and radius determine the endpoint values; the endpoint tests determine the brackets.
| Left endpoint test | Right endpoint test | Interval notation |
|---|---|---|
| Diverges at \(c-R\) | Diverges at \(c+R\) | \((c-R,c+R)\) |
| Converges at \(c-R\) | Diverges at \(c+R\) | \([c-R,c+R)\) |
| Diverges at \(c-R\) | Converges at \(c+R\) | \((c-R,c+R]\) |
| Converges at \(c-R\) | Converges at \(c+R\) | \([c-R,c+R]\) |
These four possibilities are for finite positive \(R\) and real \(x\). The notation must preserve the order of the endpoints: \(c-R\) is the left endpoint and \(c+R\) is the right endpoint. A bracket is not a statement about absolute convergence in particular; it means that the original power series converges at that endpoint. If it converges only conditionally, the endpoint is still included.
Why the Endpoint Tests Are Separate
The interior condition uses a strict inequality, \(|x-c|<R\). At either boundary point, the distance equals the radius, so the absolute-convergence result for points strictly inside does not settle the question. The endpoint substitution changes the power-series terms into a particular numerical sequence, and that sequence may have its own convergence behavior.
The Ratio Test also needs careful wording here. In a case where a coefficient-ratio limit exists and gives a finite positive radius, the corresponding ratio of term magnitudes at a boundary may equal \(1\), leaving the Ratio Test inconclusive. But a ratio limit need not exist for every power series. It is therefore not generally correct to say that the ratio at every endpoint reaches \(1\). The reliable procedure is to substitute the endpoint and test the resulting series using a suitable convergence test.
Common Mistake: Giving the Radius as the Interval
A frequent error is to find \(c\) and \(R\), write down the two boundary points, and automatically include both. For the geometric example above, this would produce the incorrect interval \([-1,5]\), as if knowing that \(R=3\) settled the endpoints.
The endpoint calculations show the correction:
Neither endpoint belongs in the interval, so the correct answer is \((-1,5)\). Another common slip is to use \(R\) itself as an endpoint rather than measuring from the center. The endpoints are \(c-R\) and \(c+R\), not simply \(-R\) and \(R\) unless the center is \(0\).
A Reliable Endpoint Checklist
For this endpoint procedure, assume the radius is finite and positive.
Calculate \(c-R\) and \(c+R\), then write the open interior \((c-R,c+R)\).
Simplify the power-series terms and determine whether the resulting series converges.
Test this resulting series independently; its behavior may differ from the left endpoint’s.
Use a square bracket for each convergent endpoint and a parenthesis for each divergent endpoint.
The distinction matters because two correct answers about the same series are often presented in different forms: the radius summarizes the distance from the center, while the interval records the exact real inputs where the series converges. For a complete description, give the interval and make sure its endpoint notation reflects the separate tests.
Check Your Understanding
For each question, distinguish what the radius tells you from what must be established by endpoint testing.
- A power series has center \(c=4\) and radius \(R=2\). What are its two boundary points and its open interior?
- If the series converges at \(c-R\) and diverges at \(c+R\), what interval notation should be used?
- Why does knowing the radius alone not determine whether either endpoint is included?
- At \(x=c+R\), a substituted series converges conditionally. Should the right endpoint be included? Explain.
- Why is it unsafe to claim that the Ratio Test always gives a ratio limit of \(1\) at every power-series endpoint?