Interval of Convergence for Taylor Series: How to Find It

The interval of convergence for a Taylor series is the set of all x-values for which the series converges to the function it represents. It is always centered at the expansion point a and extends R units in each direction, giving (a − R, a + R) before you check the endpoints. Once you know the center a and the radius of convergence R, the interval is open at both ends until endpoint testing proves otherwise. This article shows how to find R with the ratio test and how to settle the endpoint question.

A Taylor series represents a function as an infinite sum of terms built from the function’s derivatives at a single point. The center is that point, and the radius is the distance from it out to the nearest point where the series breaks down.

What the Interval Means

A Taylor series is a power series, so its behavior is governed by the same three quantities every power series has: a center, a radius, and an interval. The center is the point a where the derivatives are evaluated. The radius of convergence R is a nonnegative number—possibly zero or infinite—that measures how far from a the series is guaranteed to converge. The interval of convergence is the actual set of x-values where convergence occurs.

Convergence is always symmetric about the center: if the series converges at some x, it converges at every point closer to a. That symmetry is why the interval is built around a rather than scattered arbitrarily along the number line. For a typical series, the interval has the form (a − R, a + R), but the two endpoints require separate investigation because convergence at distance exactly R is not guaranteed by the ratio test.

Three cases cover everything. If R = 0, the series converges only at the center itself, so the interval collapses to the single point {a}. If R is a finite positive number, the interval is bounded and its endpoints decide whether it is open, closed, or half-open. If R = ∞, the series converges for every real x, so the interval is (−∞, ∞).

Radius Versus Interval

The radius R answers “how far out does convergence reach?” The interval answers “which x-values are actually included?” These are different questions, and conflating them is the most common mistake. A radius of 3 centered at 2 gives (2 − 3, 2 + 3) = (−1, 5) as a candidate interval, but you cannot yet say whether −1 and 5 belong.

Using the Ratio Test

The ratio test is the standard tool for computing R, and it works the same way for Taylor series as for any power series. Write the series in the form Σ cₙ(x − a)ⁿ, where cₙ is the coefficient of the nth term. The ratio test examines the limit of the absolute ratio of consecutive terms:

L = lim (n → ∞) |cₙ₊₁(x − a)ⁿ⁺¹ / (cₙ(x − a)ⁿ)| = |x − a| · lim |cₙ₊₁ / cₙ|.

The series converges absolutely when L < 1 and diverges when L > 1. Setting L < 1 and solving for the distance from the center produces the radius. The standard formula is R = 1 / lim |cₙ₊₁ / cₙ|, which is equivalent to the root-test version R = lim |cₙ / cₙ₊₁|. Either form gives the same answer; choose whichever simplifies faster for the coefficients you have.

Worked ratio test for eˣ. The Maclaurin series for eˣ is Σ xⁿ / n!, so cₙ = 1 / n!. Then |cₙ₊₁ / cₙ| = |(1 / (n+1)!) / (1 / n!)| = n! / (n+1)! = 1 / (n + 1). As n → ∞ this ratio tends to 0, so L = |x − 0| · 0 = 0 for every x. Since L < 1 always, R = ∞ and the series converges for all real numbers. That is why the interval of convergence for the eˣ series is (−∞, ∞) with no endpoint cases to consider.

Worked ratio test for 1/(1 − x). The geometric series Σ xⁿ has cₙ = 1, so the ratio |cₙ₊₁ / cₙ| equals 1 for every n. Then L = |x|, and L < 1 requires |x| < 1, giving R = 1. This R is finite, so the endpoints x = −1 and x = 1 need their own tests.

If the coefficient ratio does not settle into a clean limit, the root test form R = lim |cₙ / cₙ₊₁| is often easier to evaluate, since it avoids carrying the power of x through the algebra. Both are standard results covered in any calculus sequence and in references such as Paul’s Online Math Notes.

When the Ratio Test Is Inconclusive

At exactly |x − a| = R, the ratio test gives L = 1 and says nothing. That is not a failure of your algebra; it is the test’s built-in blind spot. The endpoints must always be handled by a different method.

Checking the Endpoints

Because the ratio test is silent at L = 1, you plug each endpoint into the original series and treat the result as an ordinary numerical series—now with no x left in it. Three tools do most of the work.

• If the endpoint series alternates in sign, apply the alternating series test: convergence holds if the terms decrease in magnitude and approach zero.
• If all terms are positive, try the p-series test, the comparison test, or the integral test.
• If the endpoint series is a familiar form such as a geometric series, convergence or divergence may be immediate.

The results determine the bracket style. If both endpoints converge, the interval is closed: [a − R, a + R]. If neither converges, it is open: (a − R, a + R). If exactly one converges, the interval is half-open. Treat the two endpoints as completely independent problems—never assume they behave the same way.

Endpoints for the geometric series. At x = 1, Σ 1ⁿ = 1 + 1 + 1 + … diverges because the terms do not approach zero. At x = −1, Σ (−1)ⁿ = 1 − 1 + 1 − 1 + … also diverges, since the partial sums oscillate and never settle. Neither endpoint is included, so the interval is the open interval (−1, 1). This matches the known fact that 1/(1 − x) equals the series only inside that range.

A case where endpoints differ. For Σ xⁿ / n (the series related to −ln(1 − x)), the ratio test again yields R = 1. At x = 1 the series becomes the harmonic series Σ 1/n, which diverges. At x = −1 it becomes the alternating harmonic series Σ (−1)ⁿ / n, which converges by the alternating series test. The interval is therefore [−1, 1)—closed on the left, open on the right. Students who stop after finding R = 1 miss this distinction entirely.

Power series can be differentiated and integrated term by term inside their interval of convergence, which makes Taylor series central to solving differential equations and to numerical approximation; MIT OpenCourseWare treats this radius-and-endpoint procedure in full. The single most important habit is to test endpoints separately, every time.

Worked Examples

Example 1: sin x. The Maclaurin series is Σ (−1)ⁿ x²ⁿ⁺¹ / (2n+1)!. Working with the ratio of successive terms gives a limit of 0 for every fixed x, so R = ∞ and the interval is (−∞, ∞). No endpoints exist to check because the interval is unbounded.

Example 2: 1/(1 + x²). Rewriting as Σ (−1)ⁿ x²ⁿ gives cₙ = (−1)ⁿ for the even powers and 0 for odd ones. The ratio test yields R = 1. At x = 1, the series is Σ (−1)ⁿ, which diverges; at x = −1, the same terms appear, also divergent. The interval is (−1, 1).

Example 3: ln(1 + x). The series Σ (−1)ⁿ⁺¹ xⁿ / n has cₙ = (−1)ⁿ⁺¹ / n. The ratio test gives R = 1. At x = 1 the series becomes the alternating harmonic series, which converges. At x = −1 it becomes Σ (−1)ⁿ⁺¹ (−1)ⁿ / n = Σ −1/n, the negative harmonic series, which diverges. The interval is therefore (−1, 1].

A quick general table for the most common cases:

Notice that the center can be any real number, not just zero. A series for a function expanded about a = 3 with R = 2 produces the interval (1, 5) before endpoint testing; the same R applied at a different center shifts the whole interval left or right. Always anchor your interval to the center, then extend R in both directions.

For the more delicate series that appear in physics and engineering coursework, the ratio test plus endpoint analysis remains the routine procedure. If a problem involves a function with singularities, remember that R equals the distance from the center to the nearest such point—a quick sanity check on any answer you compute.

Key Takeaways

• A Taylor series is a power series, so its interval of convergence is centered at the expansion point a and has the form (a − R, a + R) before endpoint testing.
• Find R with the ratio test: R = 1 / lim |cₙ₊₁ / cₙ|, or equivalently R = lim |cₙ / cₙ₊₁| when that limit is simpler.
• The ratio test is inconclusive at |x − a| = R, so each endpoint must be tested separately as a numerical series.
• Endpoint results dictate the brackets: both converge → closed interval; one converges → half-open; neither → open.
• Common results worth memorizing: eˣ, sin x, and cos x have R = ∞; 1/(1 − x) has interval (−1, 1); ln(1 + x) has interval (−1, 1].