Radius and Interval of Convergence

The series never promised to work there. The radius of convergence tells you exactly where the series is valid: it is the distance from the center within which the infinite sum matches the function.

The radius of convergence is a single number R. For |x| < R the series converges; for |x| > R it diverges. The interval of convergence is the full set of x where the series works, which may include one or both endpoints x = ±R. Stewart §11.8 defines this and gives the three possible cases: convergence only at the center, convergence for all x, or convergence within a finite radius with endpoints that need separate testing.

The ratio test is the standard method to find the radius of convergence for a power series Σ c_n (x − a)^n. Compute L = lim_{n→∞} |c_{n+1} / c_n|. If L = 0 the radius is infinite and the series converges for all x. If L = ∞ the radius is zero and the series converges only at the center. Otherwise R = 1/L, and the series converges for |x − a| < R.

The ratio test fails when L = 1; at that boundary you must test the endpoints individually. This is the most common error: stopping at the radius and assuming the interval is open.

Once you have the radius R from the ratio test, plug x = a + R and x = a − R into the original series. Each endpoint is a numerical series that you test with the alternating series test, the p-series test, or a comparison test. The endpoint may converge absolutely, converge conditionally, or diverge.

For Σ x^n / n with center a=0 and R=1, x=1 gives the harmonic series Σ 1/n, which diverges. x=−1 gives Σ (−1)^n / n, an alternating series that converges conditionally. So the interval is [−1, 1). For Σ x^n / n^2, both endpoints converge absolutely because Σ 1/n^2 is a convergent p-series with p=2, so the interval is [−1, 1].

The Maclaurin series for e^x is Σ x^n / n! from n=0 to ∞. The general term is c_n = 1/n!. Apply the ratio test: |c_{n+1} / c_n| = 1/(n+1). The limit as n→∞ is 0. Since L = 0, the radius of convergence is infinite: R = ∞. The series converges absolutely for every real x. This matches the entity space: factorial growth in the denominator outruns any finite x^n.

The interval of convergence is (−∞, ∞). No endpoints to check. The series equals e^x for all x.

The Maclaurin series for 1/(1−x) is Σ x^n from n=0 to ∞, the geometric series. The coefficient c_n = 1. Ratio test: |c_{n+1} / c_n| = 1/1 = 1. So L = 1, and R = 1/L = 1. The series converges for |x| < 1.

Check endpoints. At x = 1, the series is Σ 1, which diverges because terms do not approach zero. At x = −1, the series is Σ (−1)^n, which also diverges (oscillates). The interval of convergence is strictly (−1, 1). Outside this interval, the series diverges and the geometric sum formula 1/(1−x) gives no meaning to the series.

The Maclaurin series for ln(1+x) is Σ (−1)^{n+1} x^n / n from n=1 to ∞.The limit as n→∞ is 1, so R = 1. Convergence for |x| < 1.

At x = 1, the series becomes Σ (−1)^{n+1} / n, which is the alternating harmonic series. It converges conditionally by the alternating series test.The interval of convergence is (−1, 1].

For the series Σ x^n representing 1/(1−x), pick x = 1.5. The partial sums are 1, 2.5, 4.75, 8.125, and grow without bound. The graph of partial sums versus number of terms shows no asymptotic approach to a finite value; the sums explode exponentially. For x = 0.5, the partial sums approach 2, the exact value of 1/(1−0.5). For x = −0.5, the partial sums oscillate but approach 2/3.

7, after 20 terms over 3300. The series does not converge to any finite number. This is the failure case: a Maclaurin series is not a global representation of the function. It is valid only within its interval of convergence.

The interval of convergence is the set of x where the power series converges. It is centered at the expansion point a and has half-length R. The ratio test gives R, but the interval includes only those endpoints where the endpoint series converges. Stewart §11.8 lists the three possibilities: convergence only at a, convergence for all x, or convergence on an interval of length 2R.

For the Maclaurin series for sin x, the interval is (−∞, ∞). For the binomial series (1+x)^p, the interval is (−1, 1) with possible endpoint inclusion depending on p. For the Maclaurin series for arctan x, the interval is [−1, 1] with conditional convergence at both ends.

When you apply the ratio test to a power series Σ c_n (x − a)^n, you treat x as a constant for the limit. The limit L becomes a function of x only through the ratio of coefficients. Compute lim_{n→∞} |c_{n+1} / c_n|. If this limit is 0, the series converges for all x. If infinite, only at x = a. Otherwise, the radius R = 1/L.

For the series Σ n! x^n, c_n = n!. Ratio limit: |(n+1)! / n!| = n+1 → ∞, so R = 0. The series converges only at x = 0. Contrast with Σ x^n / n!, which gives R = ∞. The ratio test is the primary tool for finding the radius of convergence.

Common Radii and Endpoint Behavior

Maclaurin series convergence is determined by the same radius and interval analysis. Each common Maclaurin series has a known radius from Stewart Table 1: e^x, sin x, cos x, sinh x, cosh x all have R = ∞. The geometric series 1/(1−x), ln(1+x), arctan x, and (1+x)^p have R = 1. These radii come from the ratio test on the coefficient patterns.

The convergence behavior at endpoints varies. For ln(1+x), the alternating sign at x=1 gives conditional convergence. For arctan x, both endpoints converge conditionally. For 1/(1−x), both endpoints diverge. Memorizing these intervals from the table saves time on exams, but the ratio test method applies to any power series.

Series	Radius R	Interval of Convergence
e^x = Σ x^n / n!	∞	(−∞, ∞)
sin x = Σ (−1)^n x^(2n+1) / (2n+1)!	∞	(−∞, ∞)
cos x = Σ (−1)^n x^(2n) / (2n)!	∞	(−∞, ∞)
1/(1−x) = Σ x^n	1	(−1, 1)
ln(1+x) = Σ (−1)^(n+1) x^n / n	1	(−1, 1]
arctan x = Σ (−1)^n x^(2n+1) / (2n+1)	1	[−1, 1]
(1+x)^p = Σ (p choose n) x^n	1	(−1, 1) typically