How to Find the Interval of Convergence
Find the interval of convergence in four steps: apply the ratio test, solve |x − c| < R, test both endpoints, and write the interval. With worked examples.
How to Find Interval of Convergence: Steps
Most students mistake the radius for the interval. The radius R tells you the distance from the center where convergence is guaranteed, but it says nothing about the endpoints. To find the interval of convergence, you need four steps: apply the ratio test to the general term, solve the inequality |x − c| < R for the open interval, test each endpoint separately, and then write the final interval. Here is what each step looks like on a calculus exam.
Step 1: Apply the Ratio Test to the General Term
Write the power series in the form Σ aₙ (x − c)ⁿ. Identify the center c and the coefficient aₙ. Compute the limit L = limn→∞ |aₙ₊₁ / aₙ|. This limit depends only on the coefficients, not on x. For example, for Σ (x−2)ⁿ / n², aₙ = 1/n², and the ratio test gives L = lim n²/(n+1)² = 1. The ratio test requires L|x−c| < 1 for convergence, so the next step solves for x.
Step 2: Solve |x − c| < R
From L|x−c| < 1, the radius R = 1/L. If L = 0, then R = ∞ and the series converges for all x. If L = ∞, then R = 0 and the series converges only at x = c. Otherwise, the open interval is (c − R, c + R). For the series Σ (x−2)ⁿ / n², R = 1/1 = 1, so the open interval is (1, 3).
Step 3: Test Each Endpoint Separately
Substitute x = c + R and x = c − R into the original series. Do not reuse the ratio test, it is inconclusive at endpoints. Apply p-series, alternating series, comparison, or integral tests instead. For Σ (x−2)ⁿ / n² at x = 3, the series becomes Σ 1/n², a convergent p-series with p = 2. At x = 1, it becomes Σ (−1)ⁿ / n², which converges by the alternating series test because 1/n² decreases to zero. Both endpoints converge, so the interval becomes [1, 3].
Step 4: Write the Interval
Combine the open interval with the endpoint results. Possible forms: (c−R, c+R), [c−R, c+R), (c−R, c+R], or [c−R, c+R]. For the example above, the final interval is [1, 3]. A single point interval, R = 0, is written as {c}.
Worked Example 1: Σ xⁿ / n!
Center c = 0. aₙ = 1/n!. Ratio test: lim |aₙ₊₁ / aₙ| = lim (1/(n+1)!) / (1/n!) = lim 1/(n+1) = 0. So L = 0, R = ∞. The open interval is (−∞, ∞). No finite endpoints to test, so the interval of convergence is (−∞, ∞). This matches the Taylor series for eˣ, which converges for all real x (Stewart, Calculus, 9th ed., section 11.10).
Worked Example 2: Σ (x−2)ⁿ / n²
Center c = 2. aₙ = 1/n². Ratio test: lim |aₙ₊₁ / aₙ| = lim n²/(n+1)² = 1. So L = 1, R = 1. Open interval: (1, 3). Test x = 1: series becomes Σ (−1)ⁿ / n², converges by alternating series test (terms decrease to zero). Test x = 3: series becomes Σ 1/n², converges as p-series with p = 2 > 1. Final interval: [1, 3] (OpenStax Calculus Vol. 2, 2nd ed., section 6.1, Example 6.4).
When to Use the Root Test Instead of the Ratio Test
The ratio test works for most power series, but the root test is cleaner when aₙ contains an exponent of n. Compare the two in the decision guide below.
Decision Guide: Ratio Test vs. Root Test
Use the ratio test (lim |aₙ₊₁ / aₙ|) when aₙ involves factorials, like aₙ = 1/n!. The ratio cancels the factorial cleanly. Use the root test (lim |aₙ|1/n) when aₙ has an exponent of n, like aₙ = (n/(n+1))ⁿ, where the root test reduces to a known limit. Both tests give the same radius R when the limit exists. When the ratio test limit does not exist, use the root test with lim sup (Cauchy-Hadamard theorem). For aₙ = (2 + (−1)ⁿ)⁻ⁿ, the ratio test fails but the root test with lim sup gives R = 1/3.
Common Mistakes Checklist
- Forgetting absolute values: the ratio test always uses |aₙ₊₁ / aₙ|, not aₙ₊₁ / aₙ. Omitting absolute values gives the wrong radius.
- Stopping at the open interval: the ratio test cannot decide endpoints. Test each endpoint separately.
- Assuming symmetric convergence: one endpoint may converge while the other diverges. For the Taylor series of ln(1+x), the interval is (−1, 1], the right endpoint converges conditionally (Stewart, 11.10).
- Reapplying the ratio test at endpoints: endpoint tests require p-series, alternating series, or comparison tests. The ratio test is inconclusive at |x−c| = R.
- Algebra errors with factorials: (n+1)! = (n+1) · n!, not n! + 1. Simplify factorials before taking the limit.
- Misidentifying the center: a series centered at c = 2 still uses |x−2|, not |x|. The interval is symmetric about c, not zero.
Honest Caveat About This Subject
The standard four-step procedure works for every power series you will see on a calculus exam. But the single most common mistake, and the one that costs the most points, is skipping endpoint testing. After you find R, treat the endpoints as separate problems. Apply the correct test for each one and write the result explicitly. The ratio test and root test are tools for the open interval only. Using them at the endpoints is not just wrong; it guarantees a wrong answer.
Common Questions
What is the difference between the radius and the interval of convergence?
The radius R is a non-negative number that tells you the half-length of the interval. The interval of convergence is the set of x-values themselves. The radius alone does not tell you whether the endpoints converge.
Can the interval of convergence be a single point?
Yes. When R = 0, the series converges only at x = c. For example, Σ n! xⁿ has R = 0. The interval is {0}.
What do I do when the ratio test limit does not exist?
Use the root test with lim sup (Cauchy-Hadamard theorem). The ratio test fails when aₙ = 0 for some n or when the limit oscillates. The root test is more general.
How do I choose which convergence test to use at an endpoint?
Simplify the series after substituting x = c ± R. If it becomes a p-series (Σ 1/nᵖ), use the p-series test. If it alternates, use the alternating series test. If neither, try the comparison test or integral test.
Is the interval of convergence always symmetric about the center?
The open interval is symmetric: (c−R, c+R). But endpoint convergence can be asymmetric. One endpoint may converge while the other diverges, giving an interval like [−1, 1) for the binomial series (1+x)^(1/2).
Can the interval of convergence include endpoints that converge conditionally?
Yes. Conditional convergence occurs only at endpoints.
What happens to the interval when I differentiate or integrate a power series?
The radius R stays the same. The interval of convergence may change at the endpoints. Differentiation can lose convergence at an endpoint; integration can gain it. Always re-test endpoints after differentiation or integration.