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Interval of Convergence

Interval of Convergence

For a power series in one variable, the set of values of the variable for which the series converges. The interval of convergence may be as small as a single point or as large as the set of all real numbers.

 

Example: Series sum(x^n/n) from n=1 to infinity = x + x²/2 + x³/3 + x⁴/4 + ··· converges for −1 ≤ x < 1; interval [−1,1).

 

 

See also

Convergence tests, power series convergence, radius of convergence, Taylor series, Maclaurin series, interval notation

Key Formula

∑n=0∞an(x−c)nconverges when x∈(c−R, c+R)\sum_{n=0}^{\infty} a_n (x - c)^n \quad \text{converges when } x \in (c - R,\, c + R)
Where:
  • ana_n = The coefficient of the nth term of the power series
  • xx = The variable whose values determine convergence
  • cc = The center of the power series
  • RR = The radius of convergence, found using the Ratio Test or Root Test

Worked Example

Problem: Find the interval of convergence of the power series ∑ (x^n) / n from n = 1 to ∞.
Step 1: Apply the Ratio Test to find the radius of convergence. Compute the limit of |a_{n+1}/a_n|.
∣xn+1/(n+1)xn/n∣=∣x∣⋅nn+1→∣x∣ as n→∞\left|\frac{x^{n+1}/(n+1)}{x^n/n}\right| = |x| \cdot \frac{n}{n+1} \to |x| \text{ as } n \to \infty
Step 2: The Ratio Test says the series converges when the limit is less than 1. So convergence occurs when |x| < 1, giving a radius of convergence R = 1. The open interval is (−1, 1).
∣x∣<1  ⟹  R=1|x| < 1 \implies R = 1
Step 3: Check the left endpoint x = −1. Substituting gives the alternating harmonic series, which converges by the Alternating Series Test.
∑n=1∞(−1)nnconverges (alternating harmonic series)\sum_{n=1}^{\infty} \frac{(-1)^n}{n} \quad \text{converges (alternating harmonic series)}
Step 4: Check the right endpoint x = 1. Substituting gives the harmonic series, which diverges.
∑n=1∞1ndiverges (harmonic series)\sum_{n=1}^{\infty} \frac{1}{n} \quad \text{diverges (harmonic series)}
Step 5: Include the left endpoint and exclude the right endpoint to form the interval of convergence.
Interval of convergence: [−1,1)\text{Interval of convergence: } [-1, 1)
Answer: The interval of convergence is [−1, 1).

Another Example

This example shows an edge case where the interval of convergence is just a single point, illustrating that not every power series converges on an interval of positive length.

Problem: Find the interval of convergence of ∑ (n! · x^n) from n = 0 to ∞.
Step 1: Apply the Ratio Test. Compute the limit of the ratio of consecutive terms.
∣(n+1)!⋅xn+1n!⋅xn∣=(n+1)∣x∣\left|\frac{(n+1)! \cdot x^{n+1}}{n! \cdot x^n}\right| = (n+1)|x|
Step 2: As n → ∞, the limit (n+1)|x| → ∞ for every x ≠ 0. The Ratio Test says the series diverges whenever this limit exceeds 1.
lim⁡n→∞(n+1)∣x∣=∞for all x≠0\lim_{n \to \infty}(n+1)|x| = \infty \quad \text{for all } x \neq 0
Step 3: At x = 0, every term after the first is 0, so the series trivially converges to the value of the n = 0 term.
∑n=0∞n!⋅0n=1\sum_{n=0}^{\infty} n! \cdot 0^n = 1
Answer: The interval of convergence is the single point {0}, and the radius of convergence is R = 0.

Frequently Asked Questions

What is the difference between the interval of convergence and the radius of convergence?
The radius of convergence R is a single non-negative number (or ∞) that tells you how far from the center the series converges. The interval of convergence is the actual set of x-values, which starts as the open interval (c − R, c + R) and then may include one or both endpoints after separate testing. In short, R gives the half-width; the interval gives the precise set.
Do you always have to check the endpoints?
Yes, whenever the radius of convergence R is a finite positive number. The Ratio Test and Root Test are inconclusive at the endpoints (the limit equals exactly 1), so you must substitute each endpoint into the series and use another convergence test—such as the Alternating Series Test, p-series test, or Comparison Test—to decide inclusion.
Can the interval of convergence be all real numbers?
Yes. If R = ∞, the power series converges for every real x, so the interval of convergence is (−∞, ∞). Common examples include the Maclaurin series for e^x, sin x, and cos x. In this case there are no finite endpoints to check.

Interval of Convergence vs. Radius of Convergence

Interval of ConvergenceRadius of Convergence
What it isA set (interval) of x-valuesA single number R ≥ 0 (or ∞)
Includes endpoints?May include one, both, or neither endpointDoes not address endpoints
How to find itFind R, then test each endpoint separatelyUse the Ratio Test or Root Test on the general term
NotationAn interval like [c − R, c + R) or (c − R, c + R]A number, e.g. R = 3
Possible extreme casesSingle point {c} or all of (−∞, ∞)R = 0 or R = ∞

Why It Matters

You encounter the interval of convergence in AP Calculus BC and college calculus courses whenever you work with power series, Taylor series, or Maclaurin series. Knowing the interval tells you precisely which x-values you can safely substitute into a power series representation of a function. Using a power series outside its interval of convergence gives meaningless results, so this concept is essential for applications in physics, engineering, and numerical analysis.

Common Mistakes

Mistake: Forgetting to check the endpoints after finding the radius of convergence.
Correction: The Ratio or Root Test is inconclusive when |x − c| = R. You must substitute each endpoint into the series and apply a separate convergence test (e.g., p-series, Alternating Series Test) to determine whether each endpoint is included.
Mistake: Assuming the interval is always symmetric with both endpoints included or both excluded.
Correction: While the open interval (c − R, c + R) is always symmetric, the two endpoints can behave differently. One endpoint might converge while the other diverges, leading to a half-open interval like [c − R, c + R).

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