Testing the interval of convergence endpoints

You find the radius of convergence. You write the open interval. Then you stop, and you lose points on the exam.

The ratio test gives R = 1, but it tells you nothing about what happens at x = c ± 1. At those endpoints the ratio test limit is exactly 1, which is inconclusive. You have to test each endpoint separately using a different convergence test. Endpoint testing determines whether each boundary is open or closed.

The ratio test compares the absolute value of successive terms: lim |a_{n+1}/a_n| = L. At the endpoints of the interval, x = c ± R, this limit is exactly 1. The test is inconclusive by design. The same is true for the root test: lim |a_n|^{1/n} = 1 at the endpoints. Both tests give you the radius R, but they stop at the boundary. As Stewart explains in section 11.6, the ratio test 'says nothing' when L = 1. Find the interval by first using the ratio or root test to get R, then separately test each endpoint with tests that work at L = 1: the p-series test, the alternating series test, the comparison test, or the integral test.

After you have found R and the center c, write the two candidate endpoints: x = c + R and x = c − R. Plug each one into the original power series Σ a_n (x − c)^n. The (x − c)^n factor becomes (R)^n or (−R)^n. Simplify the general term. Often the series collapses into a recognisable form: a p-series Σ 1/n^p, an alternating p-series Σ (−1)^n / n^p, a geometric series Σ r^n, or something comparable.

For example, if the original series is Σ (x^n)/n and R = 1, then at x = 1 you get Σ 1/n, the harmonic series, which diverges. At x = −1 you get Σ (−1)^n / n, the alternating harmonic series, which converges conditionally. The divergence test (nth-term test) is your first check: if the simplified terms do not approach zero, the series diverges immediately.

Once you have simplified the endpoint series, choose a convergence test from this list. Apply them in this order:

• nth-term test: If lim of the term does not equal 0, the series diverges. Stop.
• p-series test: If the series is of the form Σ 1/n^p, it converges if p > 1 and diverges if p ≤ 1.
• Alternating series test: If the series alternates signs and the absolute values of terms decrease to 0, the series converges.
• Comparison test: Compare the series to a known convergent (p-series with p > 1) or divergent (p-series with p ≤ 1) series. Use the limit comparison test when the terms are similar but not identical.
• Integral test: Use when the terms are positive, decreasing, and can be integrated as a function. The series converges if the improper integral converges.

OpenStax Calculus Vol. 2 sections 5.3-5.6 cover the divergence test, integral test, comparison tests, and alternating series test in detail. The p-series test is a special case of the integral test but is faster to apply when the series is exactly 1/n^p.

A power series converges absolutely on the open interval (c−R, c+R). At the endpoints, convergence may be absolute or conditional. Conditional convergence means the series converges but the series of absolute values diverges. The alternating harmonic series Σ (−1)^n / n at x = 1 (for ln(1+x)) is the classic example: it converges conditionally because the harmonic series of absolute values diverges.

If the endpoint series has all positive terms (or all negative terms), it cannot be conditionally convergent. In that case, if the series converges, it converges absolutely. If the endpoint series alternates, check absolute convergence separately after the alternating series test confirms convergence. Stewart's section 11.10 discusses conditional convergence for power series endpoints.

Follow these steps in order for each endpoint:

• Step 1: Substitute x = c ± R and simplify the general term.
• Step 2: Apply the nth-term test. If the limit is not zero, the endpoint is excluded. Stop.
• Step 3: Identify the series type: is it a p-series? An alternating p-series? Something else?
• Step 4: Apply the appropriate test: p-series test for Σ 1/n^p, alternating series test for Σ (−1)^n b_n, comparison test for other forms.
• Step 5: If the alternating series test succeeds, check absolute convergence by testing Σ |term|. If the absolute series diverges, the endpoint converges conditionally.
• Step 6: Record the result: endpoint included (converges) or excluded (diverges). Repeat for the other endpoint.

Example 1: Both Endpoints Included

The series Σ x^n / n^2 has center c = 0. Using the ratio test on a_n = 1/n^2 gives lim |a_{n+1}/a_n| = 1, so R = 1. At x = 1, the series is Σ 1/n^2, a p-series with p = 2 > 1, which converges. At x = −1, the series is Σ (−1)^n / n^2, which converges absolutely by the p-series test (absolute value gives Σ 1/n^2). Both endpoints are included. The interval of convergence is [−1, 1].

Example 2: One Endpoint Included

The series Σ x^n / n has center c = 0 and R = 1. At x = 1, the series is Σ 1/n, the harmonic series, which diverges (p ≤ 1). At x = −1, the series is Σ (−1)^n / n, which converges by the alternating series test (terms decrease to 0). It converges conditionally because Σ 1/n diverges. Only the left endpoint (x = −1) is included. The interval is [−1, 1).

Example 3: Neither Endpoint Included

The geometric series Σ x^n has center c = 0 and R = 1. At x = 1, the series is Σ 1, which diverges by the nth-term test (terms do not approach 0). At x = −1, the series is Σ (−1)^n, which also diverges by the nth-term test (terms do not approach 0). Neither endpoint is included. The interval is (−1, 1).

Example 4: Both Endpoints Included, Alternating

The series Σ (−1)^n x^{2n+1} / (2n+1) is the Maclaurin series for arctan x, with center c = 0 and R = 1. At x = 1, the series is Σ (−1)^n / (2n+1), which converges by the alternating series test. At x = −1, the series becomes Σ (−1)^n (−1)^{2n+1} / (2n+1) = Σ (−1)^{3n+1} / (2n+1), which also converges by the alternating series test. Both endpoints converge conditionally. The interval is [−1, 1].

Forgetting to test endpoints is the single most common error in power series problems. Students find R, write (c−R, c+R) as the answer, and move on. The ratio test is inconclusive at endpoints, so that answer is incomplete. Another frequent mistake is assuming both endpoints behave the same way. The endpoint convergence can be asymmetric: one endpoint converges while the other diverges. The series Σ x^n / n has a closed left endpoint and an open right endpoint. The series Σ x^n has both endpoints open. Always test each endpoint separately.

Another failure mode is misapplying the alternating series test to a non-alternating series. The alternating series test requires terms that alternate in sign. If the endpoint series has all positive terms, the alternating series test does not apply. Use the p-series test or comparison test instead.

Conditional convergence at an endpoint occurs when the series converges but the series of absolute values diverges. This is possible only when the endpoint series alternates in sign. The alternating harmonic series Σ (−1)^n / n is the canonical example. At the endpoint x = 1 for the series ln(1+x) (which has Maclaurin series Σ (−1)^{n+1} x^n / n), the series becomes Σ (−1)^{n+1} / n, which converges conditionally. The interval of convergence for ln(1+x) is (−1, 1].

When you encounter an alternating series at an endpoint, first apply the alternating series test. If it passes (terms decrease to 0), then check absolute convergence by testing the series of absolute values. If the absolute series diverges, the endpoint converges conditionally. This distinction matters because absolute convergence guarantees stronger properties, like term-by-term differentiation and integration, which may not hold at conditionally convergent endpoints. OpenStax Calculus Vol. 2 section 5.5 covers this.

The p-series test is the most common tool at endpoints because many power series simplify to a p-series after substitution. A p-series is of the form Σ 1/n^p. It converges if p > 1 and diverges if p ≤ 1. The harmonic series (p = 1) is the boundary case and diverges. If your endpoint series is Σ 1/n, it diverges. If it is Σ 1/n^2, it converges. If it is Σ 1/√n = Σ 1/n^{1/2}, it diverges because p = 1/2 ≤ 1.

When the endpoint series has terms like 1/(n^2 + 1) or 1/(n ln n), use the comparison test or integral test. Compare the series to a p-series. For example, Σ 1/(n^2 + 1) converges because 1/(n^2 + 1) < 1/n^2 and Σ 1/n^2 converges (p = 2 > 1). The limit comparison test is useful when the terms are asymptotically equal to a p-series.

This subject suits Calculus II students solving homework problems who need to find the interval for a given power series and verify endpoint convergence. It also suits AP Calculus BC students preparing for the free-response section, where a power series problem with endpoint testing appears every year. Tutors who need a quick reference for the ratio/root test procedure and the standard convergence tests used at endpoints will find this useful. Self-studying learners who have a series and want to check their hand calculation against a reliable method will also benefit.

Skip this if you are looking for the interval of convergence for a Fourier series, Laplace transform, or numerical method, those are different subjects in differential equations or numerical analysis. Go to a site specific to those transforms or methods.

The single thing that most often goes wrong here is forgetting to test endpoints at all. The ratio test gives a radius, not an interval. Always test each endpoint separately.