Interval of Convergence for Taylor Series

You have a Taylor series, either a standard Maclaurin series from a table or one you derived yourself, and you need to know exactly which x-values make it converge. The answer always comes in three steps: find the radius R using the ratio test or root test, write the open interval (c−R, c+R), and then test each endpoint separately. Here are those steps, the standard series intervals, and the common pitfalls that make students lose points.

The table below lists the Maclaurin series (center c = 0) for the functions you will see most often. Every entry is verified against Stewart, Calculus (sections 11.6-11.10) and OpenStax Calculus Vol. 2 (sections 5.3-5.6 and 6.1-6.3). The interval of convergence is given in the form that includes or excludes endpoints as the tests dictate.

For each series, the general term a_n is shown as a function of n. The radius R is the number that appears in the interval bounds. Notice that for entire functions (e^x, sin x, cos x, sinh x, cosh x) the radius is infinite, so the interval is all real numbers. For functions with a singularity, like 1/(1−x) or ln(1+x), the radius equals the distance from the center to that singularity.

The Standard Maclaurin Series Table

e^x: Σ xⁿ/n!, interval (−∞, ∞), R = ∞. sin x: Σ (−1)ⁿ x²ⁿ⁺¹/(2n+1)!, interval (−∞, ∞), R = ∞. cos x: Σ (−1)ⁿ x²ⁿ/(2n)!, interval (−∞, ∞), R = ∞. 1/(1−x): Σ xⁿ, interval (−1, 1), R = 1. ln(1+x): Σ (−1)ⁿ⁺¹ xⁿ/n, interval (−1, 1], R = 1. (1+x)^p (binomial): Σ (p choose n) xⁿ, interval (−1, 1) for general p, R = 1. arctan x: Σ (−1)ⁿ x²ⁿ⁺¹/(2n+1), interval [−1, 1], R = 1. ln(1−x): −Σ xⁿ/n, interval [−1, 1), R = 1. sinh x: Σ x²ⁿ⁺¹/(2n+1)!, interval (−∞, ∞), R = ∞. cosh x: Σ x²ⁿ/(2n)!, interval (−∞, ∞), R = ∞.

The endpoint behaviour matters. For ln(1+x), the series converges at x = 1 (conditionally, by the alternating series test) but diverges at x = −1 (harmonic series). For arctan x, both endpoints converge conditionally. For 1/(1−x), both endpoints diverge because the terms do not approach zero.

Series	General Term a_n (centered at x=0)	Interval of Convergence	Radius R
e^x	x^n / n!	(-∞, ∞)	∞
sin x	(-1)^n x^(2n+1) / (2n+1)!	(-∞, ∞)	∞
cos x	(-1)^n x^(2n) / (2n)!	(-∞, ∞)	∞
1/(1-x)	x^n	(-1, 1)	1
ln(1+x)	(-1)^(n+1) x^n / n	(-1, 1]	1
arctan x	(-1)^n x^(2n+1) / (2n+1)	[-1, 1]	1
ln(1-x)	-x^n / n	[-1, 1)	1
sinh x	x^(2n+1) / (2n+1)!	(-∞, ∞)	∞
cosh x	x^(2n) / (2n)!	(-∞, ∞)	∞

When a Taylor series is centered at a point a other than 0, the interval of convergence shifts. The radius R is still found from the coefficients a_n using the ratio test or root test, but the center a moves the interval: it becomes (a−R, a+R) before endpoint testing. The ratio test says: if lim |a_n+1/a_n| = L, then R = 1/L. If L = 0, R = ∞. If L = ∞, R = 0. The root test works similarly: R = 1 / lim |a_n|^1/n.

Consider the Taylor series for e^x about c = 2. The coefficients are e²/n!, so the ratio test gives R = ∞. The interval is (−∞, ∞). For ln(1+x) about c = 1, the series converges for |x−1| < 2. Testing endpoints: at x = 3 the series becomes an alternating harmonic-like series that converges conditionally; at x = −1 it diverges. The interval is (−1, 3].

Why the Center Matters

The distance from the center to the nearest singularity determines the radius. For 1/(1−x) centered at c = 0, the singularity is at x = 1, so R = 1 and the interval is (−1, 1). Center at c = 2 moves the singularity 1 unit away, giving R = 1 and interval (1, 3). The series converges for x in (1, 3) and diverges outside.

The most common mistake is forgetting to test endpoints. Students find R and write (a−R, a+R) as the final answer. The ratio test and root test are both inconclusive when L = 1. You must plug x = a+R and x = a−R into the original series and then use a p-series test, alternating series test, comparison test, or integral test to determine convergence or divergence at each endpoint.

When you substitute an expression into a known Maclaurin series, the interval of convergence of the new series comes from the original interval after substitution. For example, the Maclaurin series for e^−x² uses the e^u series with u = −x². The original e^u series converges for all u, so e^−x² converges for all x: interval (−∞, ∞).

For a more constrained case, take the series for 1/(1−u) = Σ uⁿ, which converges when |u| < 1. Substitute u = x/3 to get 1/(1−x/3) = Σ (x/3)ⁿ. The convergence condition |x/3| < 1 gives |x| < 3, so the interval is (−3, 3) before endpoint testing. Endpoints diverge for this geometric series because the terms do not approach zero.

A trickier example: the Maclaurin series for ln(1+u) converges for −1 < u ≤ 1. Substitute u = x² to get ln(1+x²) = Σ (−1)ⁿ⁺¹ x²ⁿ/n. The condition |x²| < 1 gives |x| < 1, so the open interval is (−1, 1). Now test endpoints: at x = 1, the series becomes Σ (−1)ⁿ⁺¹/n, which converges conditionally by the alternating series test. At x = −1, the terms become Σ (−1)ⁿ⁺¹ (−1)²ⁿ/n = Σ (−1)ⁿ⁺¹/n again, so it also converges. The full interval is [−1, 1].

A power series can be differentiated or integrated term-by-term inside its interval of convergence. Crucially, the radius R stays the same after differentiation or integration. The endpoint behaviour may change, so you must test endpoints again.

Start with the geometric series 1/(1−x) = Σ xⁿ for |x| < 1. Differentiate both sides: 1/(1−x)² = Σ n xⁿ⁻¹, which still converges for |x| < 1 (R = 1). At x = 1, the differentiated series becomes Σ n, which diverges by the nth-term test. At x = −1, it becomes Σ n (−1)ⁿ⁻¹, also diverges. So the interval remains (−1, 1).

Integrate the geometric series: ∫ 1/(1−x) dx = −ln(1−x) = Σ xⁿ⁺¹/(n+1) + C. The radius is still 1. But now at x = 1, the series becomes the harmonic series, which diverges. At x = −1, it becomes an alternating harmonic series that converges conditionally. The interval of convergence for −ln(1−x) is [−1, 1).

The rule: differentiation and integration preserve R, but endpoint convergence can shift from divergence to convergence (or the reverse). Always test the endpoints of the new series.

Find the interval of convergence for the Maclaurin series of ln(1+x). The series is Σ (−1)ⁿ⁺¹ xⁿ/n. Apply the ratio test: lim |a_n+1/a_n| = lim |(−1)ⁿ⁺² xⁿ⁺¹/(n+1) / (−1)ⁿ⁺¹ xⁿ/n| = lim |x| n/(n+1) = |x|. For convergence, |x| < 1 gives R = 1. The open interval is (−1, 1).

Now test x = 1: the series becomes Σ (−1)ⁿ⁺¹/n, the alternating harmonic series. The alternating series test applies: terms decrease to 0, so it converges conditionally. Test x = −1: the series becomes Σ (−1)ⁿ⁺¹ (−1)ⁿ/n = Σ (−1)²ⁿ⁺¹/n = Σ −1/n, which is the negative harmonic series, diverges. The interval of convergence is (−1, 1].

The same method works for any Taylor series: find R, write (a−R, a+R), then test endpoints with p-series, alternating series, or comparison tests.

Because the Maclaurin series for ln(1+x) appears often, its interval of convergence deserves special attention. The series converges at x = 1 (conditionally) but not at x = −1. This is a classic example of asymmetric endpoint behaviour: one endpoint converges, the other diverges.

Why does x = 1 converge? The alternating series test requires terms that decrease in absolute value to zero. For n ≥ 1, 1/n decreases and approaches 0, so the series passes. At x = −1, the series becomes the harmonic series with all terms negative, which diverges because the harmonic series diverges.

When you shift the center, the interval moves. For the Taylor series of ln(1+x) about c = 1, the radius is 2, giving the open interval (−1, 3). Testing x = 3 gives an alternating series that converges; x = −1 gives a divergent series. The interval is (−1, 3].

The ratio test formula R = lim |a_n/a_n+1| fails when the limit does not exist or when a_n = 0 for some n. In those cases, use the root test with lim sup: R = 1 / lim sup |a_n|^1/n. This is the Cauchy, Hadamard theorem, and it always works for power series.

For example, the series Σ (x/2)ⁿ / nⁿ has a_n = 1/(nⁿ 2ⁿ). The root test gives lim |a_n|^1/n = lim 1/(n * 2) = 0, so R = ∞. The series converges for all x.

When the root test also gives L = 1 (inconclusive), you still need to test endpoints separately. No single test handles endpoints; it is always a manual step.

This subject suits calculus II students who need to find the interval of convergence for homework series and verify endpoint convergence. It also fits AP Calculus BC students preparing for free-response questions, where a power series problem with endpoint testing appears every year. Tutors can use the table and worked examples as a quick reference. Self-studying learners who have a series and want to check their hand calculation will find the standard intervals and methods here.

Skip this subject if you are looking for the interval of convergence for a Fourier series, Laplace transform, or numerical method, those belong to differential equations or numerical analysis. Also skip if you only need the radius of convergence without endpoint testing; that is a different skill.

The single thing that most often goes wrong: students stop after finding R and never test endpoints. The interval of convergence is not complete until you have checked x = a+R and x = a−R individually.