Interval of Convergence Calculator
Find the radius and interval of convergence of a power series with the ratio or root test, including endpoint checks and a step-by-step explanation.
Interval of Convergence Calculator
Calculate the interval of convergence and radius of convergence for power series. This calculator uses the ratio test and root test to determine where an infinite series converges, providing detailed step-by-step solutions.
Power Series Input
Enter the general term of your power series in the form Σ aₙ(x - c)ⁿ
Series Coefficients Pattern
Define the pattern for the coefficient aₙ
Most students assume the ratio test gives the interval of convergence. It does not. The ratio test gives the radius of convergence, a single number, R. The interval requires a second, separate step: plugging each endpoint back into the original series and applying a different convergence test. This interval of convergence calculator performs both steps automatically, but understanding what it does prevents you from writing a wrong answer on a problem set.
- Radius of Convergence (R): Half-length of the interval. Found by the ratio test or root test. Formula: R = 1 / L, where L = lim |aₙ₊₁ / aₙ| or lim |aₙ|^(1/n) (if the limit exists). If L = 0, R = ∞. If L = ∞, R = 0.
- Interval of Convergence: Set of all x-values where the power series converges. It is symmetric about the center c: (c − R, c + R) plus whichever endpoints converge when tested separately.
- Endpoint Testing: Substitute x = c + R and x = c − R into the original series. Apply the p-series test, alternating series test, comparison test, or integral test. The ratio test is inconclusive at endpoints.
- Power Series Form: Σ aₙ (x − c)ⁿ, where aₙ are coefficients, c is the center, and x is the variable. Every Taylor series is a power series, but not every power series is a Taylor series.
- Standard Maclaurin Series Examples: eˣ (R = ∞), sin x (R = ∞), cos x (R = ∞), 1/(1−x) (R = 1, interval (−1, 1)), ln(1+x) (R = 1, interval (−1, 1]), arctan x (R = 1, interval [−1, 1]).
How to Enter a Series Into the Power Series Calculator
Select the series type first: General Power Series Centered at Point a, or Taylor/Maclaurin Series. Enter the center point c. Then define the coefficient pattern for aₙ. The coefficient type menu offers Constant (aₙ = k), Factorial (aₙ = n!), Power (aₙ = nᵖ), Exponential (aₙ = kⁿ), Rational (aₙ = p(n)/q(n)), Alternating (aₙ = (−1)ⁿ × f(n)), or a Custom Formula. For a custom formula, use n, numbers, +, −, *, /, ^, and parentheses; factorial notation n! or (n+1)! is recognised. The calculator compiles your formula into a numerical function and evaluates it at large n to estimate the radius.
If you enter a custom expression that contains anything other than those allowed symbols, the calculator returns an error message rather than silently assuming aₙ = 1. This is important: a typo in the coefficient formula used to produce a radius of 1 without warning, which made it look like the series converged when it did not. The error message tells you to fix the expression.
Choosing Between Ratio Test and Root Test
Pick the ratio test for factorial terms (n!), because the ratio of consecutive factorials simplifies to 1/(n+1).The automatic option uses the ratio test by default because it works for most homework problems. Both tests give the same radius when the limit exists; the root test is more general because it uses lim sup and works when the ratio test limit does not exist. OpenStax Calculus Vol. 2 section 5.5 covers both methods.
What the Calculator Does: Ratio/Root Test → R → Endpoints
The calculator fits the logarithm of the coefficients to the model ln|aₙ| ≈ ℓ·n + p·ln n + c. The value of ℓ gives L = e^ℓ, and R = 1/L. The coefficient p, the exponent of n in the terms, determines what happens at the endpoints. If p > −1, the terms do not approach zero and the series diverges at both endpoints. If p < −1, the series converges absolutely at both endpoints. If −1 ≤ p < 0 and the sign pattern alternates, the endpoint series converges conditionally. If the sign pattern is constant in that range, the endpoint series diverges.
This method avoids overflow errors that crash straightforward ratio-test code when n reaches 10⁶. The calculator evaluates the logarithm of the coefficient analytically for each supported type (factorial, power, exponential) rather than multiplying huge numbers. For factorial terms, it uses Stirling's approximation when n exceeds 50. The result panel shows the interval, radius, center, the test used, and a visual representation of the convergence line with open or closed endpoint markers.
Reading the Endpoint Results
Endpoint analysis appears in a separate row: left endpoint result and right endpoint result, each labelled "Converges absolutely ✓", "Converges conditionally ✓", "Diverges ✗", or "Inconclusive ?". The test used at each endpoint is explained in plain English, for example, "Converges conditionally: alternating series test (|terms| ~ C/n^0.5 decrease to 0), but Σ|terms| diverges like the p-series with exponent 0.5 ≤ 1."
| Step | Action | Result |
|---|---|---|
| 1. Enter the series | Series: Σ (−1)ⁿ xⁿ / n. Center c = 0. Coefficient type: Alternating, function f(n) = n. (This is the Maclaurin series for ln(1+x).) | — |
| 2. Ratio test | Calculator computes L = lim |aₙ₊₁ / aₙ| = lim n/(n+1) = 1. R = 1/1 = 1. | R = 1 |
| 3. Open interval | The series converges for |x| < 1, i.e. (−1, 1). | (−1, 1) |
| 4. Right endpoint x = 1 | Substitute x = 1: Σ (−1)ⁿ (1)ⁿ / n = Σ (−1)ⁿ / n. Alternating harmonic series. Alternating series test: terms decrease to 0. Converges conditionally. | Endpoint converges |
| 5. Left endpoint x = −1 | Substitute x = −1: Σ (−1)ⁿ (−1)ⁿ / n = Σ (1)ⁿ / n = Σ 1/n. Harmonic series, diverges. | Endpoint diverges |
| 6. Final interval | Include the converging endpoint, exclude the diverging one. | (−1, 1] |
Limits of Automatic Endpoint Testing
The calculator determines endpoint convergence by fitting the growth exponent p of the coefficient series. This works when the coefficient follows a clear pattern, constant, factorial, power, exponential, rational, alternating with one of these, or a custom formula that evaluates to a smooth function of n. When the coefficient pattern is irregular or contains a mix of growth rates that cancel at large n, the fit may produce a misleading p. In those cases, the endpoint result panel shows "Inconclusive ?" and recommends testing that endpoint by hand.
If the series is not a power series, for example, a Fourier series or a series in terms of sine and cosine, do not use this calculator. The interval of convergence concept does not apply to those series in the same way. The calculator also cannot handle series where the coefficients depend on x in a non-power way, such as a series of the form Σ aₙ (sin x)ⁿ. The input expects a clear power series centred at a constant c.
Another limit: the calculator assumes the coefficients are eventually non-zero for the ratio test method. If aₙ = 0 for infinitely many n in a pattern that breaks the ratio, the root test (lim sup version) is the correct general method, but the automatic fit may still produce a sensible answer because it evaluates the logarithm at large n where later terms dominate.
What Often Goes Wrong
The single most common mistake is taking the open interval (c − R, c + R) as the final answer without testing endpoints. The ratio test cannot decide at the endpoints, so leaving them untested is leaving the problem unfinished. The calculator tests endpoints automatically, but if you are working by hand, you must substitute each endpoint into the original series and apply the p-series test, alternating series test, comparison test, or integral test. A power series that converges at one endpoint but not the other is common, the Maclaurin series for ln(1+x) has interval (−1, 1], and the series for arctan x has interval [−1, 1]. Never assume symmetry of endpoint convergence.
Common Questions
What is the difference between the radius of convergence and the interval of convergence?
The radius of convergence R is a single number: the distance from the centre within which the series converges. The interval of convergence is the set of x-values themselves. You get the interval by starting with (c − R, c + R) and then adding whichever endpoints converge. The radius is one step in finding the interval, not the answer itself. Stewart, Calculus, section 11.8 makes this distinction explicit.
Why can two endpoints behave differently? One converges, the other does not.
The ratio test and root test are inconclusive at the boundaries, so each endpoint must be tested using a different method. At one endpoint, substitution may turn the series into an alternating series that converges conditionally (e.g., the alternating harmonic series). At the other endpoint, substitution may produce a divergent p-series (e.g., the harmonic series). The power series is symmetric about c in length, but the convergence behaviour at the two endpoints can differ because the sign pattern of the terms changes. This is covered in OpenStax Calculus Vol. 2, section 5.5.
What does it mean when R = 0 or R = ∞?
R = 0 means the series converges only at the centre x = c. The interval is just {c}. This happens when the coefficients grow faster than any exponential, for example aₙ = n!. R = ∞ means the series converges for all real numbers. This happens when the coefficients shrink faster than any exponential, for example aₙ = 1/n!. The interval is (−∞, ∞). Both cases are valid and appear in standard series such as the exponential series (R = ∞) and the series for the Bessel function of the first kind of order zero evaluated at zero (R = 0? no, that one has R = ∞ too; an example of R = 0 is Σ n! xⁿ, which converges only at x = 0).
How do I handle a series with only even powers, like Σ x^(2n)?
A power series with only even powers, such as Σ aₙ x^(2n), is still a power series in the variable (x²), but the centre and radius must be handled carefully. Write it as Σ aₙ (x²)ⁿ and treat x² as the variable for the ratio test. The radius of convergence in terms of x² is R² = 1/L, so the radius in terms of x is √(R²) = √(1/L). The interval is then (−R, R) where R is the radius in x, not in x². For example, the Maclaurin series for cos x is Σ (−1)ⁿ x^(2n)/(2n)!. The ratio test on the coefficients 1/(2n)! gives L = 0, so the radius in x² is ∞, meaning the radius in x is also ∞. The interval is (−∞, ∞).
What do I do when the interval of convergence calculator shows a misleading default like R = ∞ before I enter anything?
That default is a placeholder, not a result. The calculator displays "R = ∞" and the interval "(−∞, ∞)" in the results panel before any input is provided. This is not a judgement that the series converges everywhere, it is the initial empty state. Always enter your actual series and click "Calculate Convergence" to see the real radius and interval. If you see these values after entering a specific series, double-check that the series coefficient pattern and centre are correct. The default does not reflect any real series.
