From Radius and Endpoint Tests to an Interval
A power series’ radius of convergence tells you where to find its possible boundary points. It does not, by itself, tell you whether those boundary points belong in the convergence set. That final decision comes from testing each endpoint separately.
Writing the answer correctly means combining two pieces of information: the open interval guaranteed by the radius, and the results of the endpoint tests. The endpoints can behave differently, so the final interval need not be symmetric in its inclusion of endpoints.
The Basic Translation Rule
For a power series centered at \(c\) with finite radius \(R>0\), the series converges absolutely whenever \(|x-c|<R\) and diverges whenever \(|x-c|>R\). Thus, before testing endpoints, the certain interior is described by one inequality:
The left endpoint is \(c-R\); the right endpoint is \(c+R\). At either endpoint, the strict inequality becomes equality, so the radius result alone makes no decision. Use the result of the corresponding endpoint test to select the bracket.
| Endpoint test result | Notation at that endpoint |
|---|---|
| The series converges at the endpoint | Include the endpoint with a square bracket |
| The series diverges at the endpoint | Exclude the endpoint with a parenthesis |
Apply that rule independently on the left and right. For instance, if the left endpoint converges and the right endpoint diverges, the interval is \([c-R,c+R)\). If the left endpoint diverges and the right endpoint converges, it is \((c-R,c+R]\).
Why the Rule Works
The radius result separates the real line into three parts: points strictly inside the radius, the two boundary points, and points strictly outside. The inside points converge and the outside points diverge. Only the boundary points need individual decisions.
To see why the endpoints are the ones stated, solve the inequality \(|x-c|<R\). For real \(x\), this is equivalent to \(-R<x-c<R\). Adding \(c\) throughout gives \(c-R<x<c+R\). The interval’s left and right endpoints are therefore \(c-R\) and \(c+R\), respectively. Replacing a strict endpoint with a square bracket is justified only when the series also converges at that endpoint.
Record \(c\) and \(R\), keeping the center’s sign and the radius positive.
Compute \(c-R\) for the left endpoint and \(c+R\) for the right endpoint.
Include a boundary value if and only if the series converges there.
Use the left endpoint first and the right endpoint second; verify each bracket against its test result.
Worked Examples
Worked Example: Only the Right Endpoint Converges
Write the convergence interval for \(\sum_{n=1}^{\infty}\frac{(-1)^{n+1}(x-2)^n}{n}\), given its radius and endpoint test results.
The center is \(c=2\), and the radius is \(R=1\). The left endpoint is \(2-1=1\), and the right endpoint is \(2+1=3\). For \(1<x<3\), the series converges absolutely by the radius result; for \(x<1\) or \(x>3\), it diverges.
At \(x=1\), substitute into the general term:
The endpoint series is \(-\sum_{n=1}^{\infty}\frac1n\), which diverges, so exclude \(1\). At \(x=3\), the terms become \(\frac{(-1)^{n+1}}{n}\). This alternating series converges: its term magnitudes \(1/n\) decrease to \(0\). Include \(3\).
Therefore the convergence interval is \((1,3]\). Check: its left endpoint is parenthesized because the series diverges at \(1\), and its right endpoint has a square bracket because the series converges at \(3\).
Worked Example: Only the Left Endpoint Converges
Find the convergence interval for \(\sum_{n=1}^{\infty}\frac{(x-2)^n}{n\,3^n}\).
Write the coefficient of \((x-2)^n\) as \(a_n=\frac{1}{n3^n}\). The ratio of consecutive coefficient magnitudes is
Thus the radius is \(R=3\), with center \(c=2\). The endpoints are \(2-3=-1\) and \(2+3=5\).
At \(x=-1\), each term becomes
This alternating harmonic series converges, so include \(-1\). At \(x=5\), each term is
The harmonic series diverges, so exclude \(5\). The convergence interval is \([-1,5)\). Check: the left bracket matches convergence at \(-1\), while the right parenthesis matches divergence at \(5\).
Worked Example: Infinite Radius
Write the convergence set for \(\sum_{n=0}^{\infty}\frac{(x+1)^n}{2^n n!}\).
The center is \(c=-1\). For any fixed real \(x\), the ratio of the absolute values of consecutive terms, once the terms are nonzero, is
Since the limit is less than \(1\) for every fixed real \(x\), the ratio test gives convergence for every real \(x\); equivalently, the radius is \(R=\infty\). There are no finite endpoints to test. The convergence interval is \((-\infty,\infty)\), also written \(\mathbb{R}\). Check: every real number is included, and both infinity symbols have parentheses because infinity is not a real endpoint.
Finite and Infinite Radii
For a finite positive radius, both possible endpoints are finite, so the convergence set is bounded. The two endpoint tests can still produce four outcomes:
| Left endpoint | Right endpoint | Convergence interval |
|---|---|---|
| Diverges | Diverges | \((c-R,c+R)\) |
| Converges | Diverges | \([c-R,c+R)\) |
| Diverges | Converges | \((c-R,c+R]\) |
| Converges | Converges | \([c-R,c+R]\) |
If \(R=\infty\), the power series converges for every real \(x\), so the interval is \((-\infty,\infty)\). In standard real power series, a one-sided unbounded interval such as \([a,\infty)\) cannot be the convergence set: an infinite radius gives the whole real line, while a finite radius gives two finite boundary points and convergence only between them, subject to endpoint tests. When an interval does have an infinite bound, infinity always takes a parenthesis, never a square bracket.
If \(R=0\), the radius guarantees convergence only at the center, where every positive-power term is zero and the series reduces to its constant term. At other points, the radius result gives divergence. The convergence set is the singleton \(\{c\}\), which can be written as the degenerate closed interval \([c,c]\). Do not confuse a zero radius with an empty convergence set.
A Common Notation Mistake
A frequent error is to assign brackets to both endpoints automatically, as if the radius included its boundary. For the series in the first worked example, that would produce the incorrect answer \([1,3]\). But substitution at \(x=1\) gives \(-1/n\), and the corresponding harmonic series diverges. The corrected interval is \((1,3]\).
Another error is reversing the endpoints when the center is negative. For example, if \(c=-2\) and \(R=4\), then \(c-R=-6\) is the left endpoint and \(c+R=2\) is the right endpoint. The order follows from the calculations, not from dropping or rearranging signs by inspection.
Check Your Understanding
For each question, use the stated radius and endpoint results to write the convergence interval.
- A power series has center \(4\), radius \(2\), converges at the left endpoint, and diverges at the right endpoint. What is its convergence interval?
- A power series has center \(-3\), radius \(5\), and diverges at both endpoints. Write its convergence interval.
- A power series has infinite radius. What is its convergence interval, and how should infinity be marked?
- Can a standard power series have convergence set \((a,\infty)\)? Explain using the radius.
- A power series has center \(1\) and radius \(0\). What is its convergence set if it converges at its center?
