Interval of Convergence Calculator
For a power series ∑ a(n)·(x−c)n, enter the coefficient a(n) as a formula in n, and the center c. This calculator applies the ratio test to find the radius of convergence, then automatically tests both endpoints using the divergence test, a p-series comparison, and the alternating series test to give you the complete interval of convergence.
The open interval is always (c−R, c+R) — each endpoint, x = c−R and x = c+R, needs a separate test because the ratio test gives no answer exactly there.
Endpoint results are computed numerically (divergence test, p-series comparison, alternating series test) by sampling the series at large n. For unusual or borderline series, verify the endpoint tests by hand.
What Is the Interval of Convergence?
A power series has the form ∑n=0∞ a(n)(x−c)n — an infinite sum where each term is a coefficient a(n) multiplied by a power of (x−c). For most values of x, this infinite sum either converges to a specific finite number or diverges to infinity/oscillates. The interval of convergence is the set of all x-values where the series converges, and it is always an interval centered at c.
- Every power series converges at its center, x = c, trivially (the sum equals a(0))
- The interval of convergence is always centered at c and symmetric in width around it
- Its half-width is called the radius of convergence, R
- Inside (c−R, c+R) the series always converges; outside it always diverges
- At the two endpoints exactly, x = c−R and x = c+R, convergence must be checked separately
Finding the Radius with the Ratio Test
The series converges when L·|x−c| < 1, so:
R = 1/L (if 0 < L < ∞) | R = ∞ (if L = 0) | R = 0 (if L = ∞)
The ratio test compares each term of the series to the one before it. If the ratio of consecutive coefficients settles down to a limit L as n grows, the series behaves like a geometric series with ratio L·|x−c|, which converges exactly when that ratio is less than 1. This calculator estimates L numerically by evaluating a(n) at very large values of n (computed in log-space to safely handle factorials and exponentials without overflowing).
Why Endpoints Need Separate Tests
The ratio test's inequality, L·|x−c| < 1, is a strict inequality. At x = c+R or x = c−R, this becomes L·|x−c| = 1 exactly — the boundary case the ratio test cannot resolve either way. Substituting each endpoint back into the original series turns the power series into an ordinary numerical series (no more x), which must then be tested on its own using standard convergence tests.
The Three Endpoint Tests
| Test | What It Checks | Conclusion |
|---|---|---|
| Divergence Test (nth-term test) | Does a(n) → 0 as n → ∞? | If not, the series diverges — no further test needed |
| p-series Comparison | Does |a(n)| decay like 1/np? | Converges if p > 1; diverges if p ≤ 1 (for positive-term series) |
| Alternating Series Test | Do the terms alternate sign and shrink to 0 monotonically? | If yes, the series converges (often only conditionally) |
This calculator applies these three tests automatically at each endpoint: it first checks whether the terms shrink toward zero, then estimates the polynomial decay rate to apply a p-series-style comparison, and checks the sign pattern to detect alternating series.
The Four Possible Intervals
- Converges only at x = c — when R = 0 (radius collapses to a point)
- Open interval (c−R, c+R) — both endpoints diverge
- Half-open interval, [c−R, c+R) or (c−R, c+R] — exactly one endpoint converges
- Closed interval [c−R, c+R] — both endpoints converge
Common Series and Their Intervals
| Series | a(n) | Radius R | Interval of Convergence |
|---|---|---|---|
| ∑ xⁿ (geometric) | 1 | 1 | (−1, 1) |
| ∑ xⁿ/n! (= eˣ) | 1/n! | ∞ | (−∞, ∞) |
| ∑ n!·xⁿ | n! | 0 | {0} only |
| ∑ (−1)ⁿxⁿ/n | (−1)ⁿ/n | 1 | (−1, 1] |
| ∑ xⁿ/n² | 1/n² | 1 | [−1, 1] |
| ∑ xⁿ/(n·3ⁿ) | 1/(n·3ⁿ) | 3 | [−3, 3) |
Worked Example
Consider ∑n=1∞ (x−2)n / (n·3n), so a(n) = 1/(n·3n) and c = 2.
|a(n+1)/a(n)| = [n·3ⁿ] / [(n+1)·3ⁿ⁺¹] = n / (3(n+1)) → 1/3 as n→∞
So L = 1/3, and R = 1/L = 3
STEP 2 — Open Interval
(c−R, c+R) = (−1, 5)
STEP 3 — Endpoints
At x = 5: ∑ 1/n — the harmonic series, diverges (p-series, p=1)
At x = −1: ∑ (−1)ⁿ/n — alternating, terms → 0 monotonically, converges (alternating series test)
FINAL ANSWER Interval of convergence: [−1, 5)
Try entering a(n) = 1/(n*3^n) with center 2 into the calculator above to see this exact result.
Glossary
- Power Series
- An infinite series of the form ∑ a(n)(x−c)ⁿ, where a(n) are fixed coefficients and c is the center.
- Radius of Convergence (R)
- The distance from the center c within which the power series is guaranteed to converge; outside this distance, it always diverges.
- Ratio Test
- A convergence test that examines the limit of the ratio of consecutive terms to determine absolute convergence.
- Absolute Convergence
- A series converges absolutely when the series of absolute values of its terms also converges — a stronger condition than ordinary convergence.
- Conditional Convergence
- A series converges conditionally when it converges, but the series of absolute values of its terms diverges — common at one endpoint of an interval of convergence.
- p-series
- A series of the form ∑ 1/nᵖ, which converges if and only if p > 1.
Frequently Asked Questions
Q: What is the interval of convergence of a power series?
A: The set of all x-values for which the power series converges — always an interval centered at c with half-width equal to the radius of convergence R.
Q: How do you find the radius of convergence using the ratio test?
A: Compute L = lim |a(n+1)/a(n)|; then R = 1/L (or R = ∞ if L = 0, R = 0 if L = ∞).
Q: Why do you have to check the endpoints separately?
A: Because the ratio test's strict inequality gives no information exactly at L·|x−c| = 1, which is what happens at each endpoint.
Q: What is the divergence test (nth-term test)?
A: If the terms of a series don't approach zero, the series diverges — it's a quick first check, but it can never prove convergence.
Q: What is the alternating series test used for at endpoints?
A: It shows an alternating series converges if its terms shrink to zero monotonically — often explaining why one endpoint converges while the other diverges.
Q: What are the four possible types of interval of convergence?
A: A single point, an open interval, a half-open interval, or a closed interval — depending on the two endpoint test results.