Interval of Convergence: Worked Examples
Solved interval of convergence problems by type: R = 0, R = ∞, open, closed and half-open intervals, x^(2n) series and series centered away from zero.
Find The Interval Of Convergence For Any Power Series
You have a power series and need the exact set of x-values where it converges. The standard procedure: ratio test, then solve the inequality, then test each endpoint separately. It works for every series of the form Σ aₙ (x−c)ⁿ. Below are interval of convergence examples that cover every possible outcome: radius infinite, radius zero, open interval, half-open, closed, and series centered away from zero. Each example is fully worked with the convergence test used at each step.
Start with the ratio test: compute L = lim |aₙ₊₁ / aₙ|. Set L < 1 and solve for |x−c| < R. That gives the radius R. The open interval (c−R, c+R) is guaranteed convergence. Then plug x = c+R and x = c−R into the original series and test each endpoint separately using a p-series, alternating series test, or comparison test. The ratio test is inconclusive at endpoints, you must test them.
Example: R = ∞ (Factorial In Denominator)
Series: Σ from n=0 to ∞ of xⁿ / n!
Apply the ratio test. aₙ = 1 / n!, so aₙ₊₁ = 1 / (n+1)!.
L = lim |aₙ₊₁ / aₙ| = lim (1/(n+1)!) / (1/n!) = lim n! / (n+1)! = lim 1 / (n+1) = 0.
Since L = 0 < 1 for every x, the series converges for all real x. The radius R = ∞, and the interval of convergence is (−∞, ∞). Endpoint testing is unnecessary because there are no endpoints.
This matches the known Maclaurin series for eˣ, which converges everywhere. Stewart, Calculus, Section 11.10 lists the standard Maclaurin series for eˣ with R = ∞.
Example: R = 0 (Factorial In Numerator)
Series: Σ from n=0 to ∞ of n! xⁿ
Apply the ratio test. aₙ = n!, so aₙ₊₁ = (n+1)!.
L = lim |aₙ₊₁ / aₙ| = lim (n+1)! / n! = lim (n+1) = ∞.
Since L = ∞ > 1 for any x ≠ 0, the series converges only at x = 0. The radius R = 0. The interval of convergence is the single point {0}. No endpoint testing is needed because the only point is the center. OpenStax Calculus Vol. 2, Theorem 5.6 states that a power series can converge at exactly one point when R = 0.
Example: Open Interval (Both Endpoints Diverge)
Series: Σ from n=1 to ∞ of n xⁿ
Apply the ratio test. aₙ = n, aₙ₊₁ = n+1. L = lim (n+1)/n = 1. R = 1. Open interval (−1, 1).
Test x = 1
Plug x = 1: Σ n. Terms do not approach zero, so it diverges by the nth-term test.
Test x = −1
Plug x = −1: Σ n (−1)ⁿ. Terms do not approach zero, so it diverges by the nth-term test. Both endpoints diverge. The interval of convergence is (−1, 1), an open interval.
Example: Half-Open Interval (One Endpoint Converges)
Series: Σ from n=1 to ∞ of xⁿ / n
Apply the ratio test: aₙ = 1/n, so aₙ₊₁ = 1/(n+1). L = lim n/(n+1) = 1. R = 1. Open interval (−1, 1).
Test x = 1
Plug x = 1: Σ 1 / n. This is the harmonic series (p = 1), which diverges.
Test x = −1
Plug x = −1: Σ (−1)ⁿ / n. This is the alternating harmonic series. The terms 1/n decrease and approach zero, so the series converges by the alternating series test. The convergence is conditional, the absolute value series diverges.
The interval of convergence is [−1, 1): closed at the left endpoint, open at the right. Stewart, Calculus, Section 11.10 gives the same interval for the Maclaurin series of ln(1+x).
Example: Closed Interval (Both Endpoints Converge)
Series: Σ from n=1 to ∞ of xⁿ / n²
Apply the ratio test: aₙ = 1/n², aₙ₊₁ = 1/(n+1)². L = lim n²/(n+1)² = 1. R = 1. Open interval (−1, 1).
Test x = 1
Plug x = 1: Σ 1 / n². This is a p-series with p = 2 > 1, so it converges absolutely.
Test x = −1
Plug x = −1: Σ (−1)ⁿ / n². The absolute value series is Σ 1 / n², which converges (p = 2). So the series converges absolutely at x = −1.
Both endpoints converge. The interval of convergence is [−1, 1], a closed interval. OpenStax Calculus Vol. 2, Section 5.5 notes that the p-series test determines endpoint convergence for series of this type.
Example: X²ⁿ And Centered At C ≠ 0
Series: Σ from n=0 to ∞ of (x−2)²ⁿ / 4ⁿ
Rewrite the series as Σ ((x−2)² / 4)ⁿ. This is a geometric series with ratio r = (x−2)² / 4. A geometric series converges when |r| < 1.
Set |(x−2)² / 4| < 1 → (x−2)² < 4 → |x−2| < 2. The center is c = 2, radius R = 2. The open interval is (0, 4).
Test x = 4
Plug x = 4: (4−2)² / 4 = 4 / 4 = 1. The series becomes Σ 1ⁿ, which diverges by the nth-term test (terms do not approach zero).
Test x = 0
Plug x = 0: (0−2)² / 4 = 4 / 4 = 1. Same series, same divergence.
Both endpoints diverge. The interval of convergence is (0, 4), an open interval centered at 2.
Interval Of Convergence Practice Problems: Quick Reference Table
The table below summarizes the answers for every example. Use it to check your own interval of convergence practice problems. Each row gives the series, its center c, radius R, and the final interval of convergence.
| Series | Center c | Radius R | Interval |
|---|---|---|---|
| Σ xⁿ / n! | 0 | ∞ | (−∞, ∞) |
| Σ n! xⁿ | 0 | 0 | {0} |
| Σ n xⁿ | 0 | 1 | (−1, 1) |
| Σ xⁿ / n | 0 | 1 | [−1, 1) |
| Σ xⁿ / n² | 0 | 1 | [−1, 1] |
| Σ (x−2)²ⁿ / 4ⁿ | 2 | 2 | (0, 4) |
Common Questions
What is the difference between the radius of convergence and the interval of convergence?
The radius R is a single number, half the length of the convergence interval. The interval is the actual set of x-values, which includes endpoints only if they pass separate tests. The radius is a distance from the center, while the interval is the complete answer. OpenStax Calculus Vol. 2, Section 6.1 distinguishes them clearly.
Why is the ratio test inconclusive at the endpoints?
At an endpoint, |x−c| = R, so L = 1 exactly. The ratio test requires L < 1 for convergence and L > 1 for divergence. When L = 1, the test cannot decide. Plug the endpoint value into the original series and apply a different test, p-series, alternating series, or comparison.
When should I use the root test instead of the ratio test?
Use the root test when the coefficients aₙ involve exponents of n (like nⁿ) or when the ratio test limit is messy.Both tests give the same radius when they work, but the root test is more general, it uses lim sup and works when the ratio limit does not exist.
Can a power series converge conditionally at an endpoint?
Yes. For example, Σ (−1)ⁿ xⁿ / n converges at x = 1 (alternating harmonic series) but the absolute value series Σ 1/n diverges. That is conditional convergence. At the other endpoint x = −1, the series diverges. Conditional convergence only occurs at endpoints, never inside the open interval, where convergence is always absolute.
What does it mean when R = 0?
R = 0 means the series converges only at its center x = c. The series Σ n! xⁿ is a standard example. For any x ≠ 0, the terms grow so fast that the series diverges. This is rare but important, it shows that not all power series have a nontrivial interval.
How do I test endpoints for a series centered at c ≠ 0?
Plug x = c + R and x = c − R into the original series. Simplify the resulting numerical series. Then test each using standard convergence tests: p-series test if it simplifies to Σ 1/nᵖ, alternating series test if signs alternate, or comparison test if it resembles a known series. Treat each endpoint independently, one may converge while the other diverges.
Do I need to test endpoints if the series is geometric?
Yes. The geometric series Σ rⁿ converges only when |r| < 1. At r = ±1, it diverges. For example, Σ ((x−2)²/4)ⁿ converges for (x−2)² < 4, but at the endpoints x = 0 and x = 4, r = 1 and the series diverges. Always test endpoints even for geometric series, the ratio test is just a tool to find R, not the final answer.