Power series and the interval of convergence

When you reach for it

A polynomial c_0 + c_1 x + \cdots + c_N x^N has a value at every x. Letting the degree run on forever gives a power series centered at a,

with fixed numbers c_n, the coefficients, and the convention (x-a)^0 = 1 for every x, including x = a. For each fixed x this is an ordinary numerical series, which may converge or diverge. Where it converges, its sum depends on x, so the series defines a function f(x) = \sum c_n (x-a)^n whose domain is the set of x where the series converges. A power series is the main example of a series of functions \sum f_n(x), a series whose terms are functions of x; here f_n(x) = c_n (x-a)^n.

The geometric series is the first example: \sum_{n=0}^{\infty} x^n converges exactly when \lvert x \rvert < 1, and then its sum is \displaystyle \frac{1}{1-x}. The formula \displaystyle \frac{1}{1-x} has a value at x = 3, namely \displaystyle -\frac12, but the series at x = 3 diverges, so the series and the formula are the same function only on (-1, 1).

You reach for this lesson whenever a series carries a variable. Before a power series can be differentiated, integrated, or used to compute values of a function, you must know for which x it means anything. The answer is always an interval centered at a, found in two stages: the The ratio and root tests gives the inside of the interval, and the two endpoints are tested one at a time.

The procedure

Day 1: the three possibilities, and the radius

Everything rests on one fact about a single point.

Convergence spreads inward. If \sum c_n (x_0 - a)^n converges for some x_0 \ne a, then \sum c_n (x-a)^n converges absolutely for every x with \lvert x - a \rvert < \lvert x_0 - a \rvert.

Proof. The series converges at x_0, so its terms tend to 0 (The nth term test for divergence), and a sequence that tends to 0 is bounded: there is a number M with \lvert c_n (x_0-a)^n \rvert \le M for every n. Put \displaystyle r = \frac{\lvert x-a \rvert}{\lvert x_0-a \rvert}, which is less than 1. Then

and \sum M r^n is a convergent geometric series, so \sum \lvert c_n (x-a)^n \rvert converges by direct comparison. The contrapositive is just as useful: if the series diverges at x_1, it diverges at every x farther from a than x_1, since convergence at such an x would force convergence at x_1.

The three possibilities. For a power series \sum c_n (x-a)^n, exactly one of these holds.

• It converges only at x = a. Its radius of convergence is R = 0.
• It converges absolutely for every real x. Its radius is R = \infty.
• There is a number R > 0 such that the series converges absolutely when \lvert x-a \rvert < R and diverges when \lvert x-a \rvert > R. At the two endpoints x = a - R and x = a + R it may converge or diverge.

Proof sketch. At x = a every term after the first is 0, so a power series always converges at its center. Suppose neither of the first two cases holds: the series converges at some x_0 \ne a and diverges at some x_1. By the contrapositive above, every point of convergence lies within \lvert x_1 - a \rvert of a, so the distances from a to points of convergence are bounded. Let R be their least upper bound; that a bounded set of real numbers has a least upper bound is the completeness of the real numbers, stated here without proof. Then R \ge \lvert x_0 - a \rvert > 0. If \lvert x-a \rvert < R, some point of convergence lies farther from a than x does, and convergence spreads inward to x. If \lvert x-a \rvert > R, convergence at x would contradict the choice of R.

The set of all x where the series converges is its interval of convergence. In the third case it runs from a - R to a + R, with each end either included or not. Figure 1 shows the three cases.

Finding R with the ratio test. Fix x \ne a and apply the The ratio and root tests to the terms c_n(x-a)^n. If c_n \ne 0 and \displaystyle \left\lvert \frac{c_{n+1}}{c_n} \right\rvert \to L, then

so the series converges absolutely when L\lvert x-a \rvert < 1 and diverges when L\lvert x-a \rvert > 1. That gives \displaystyle R = \frac1L, read as R = \infty when L = 0 and R = 0 when L = \infty. The root test does the same with \displaystyle \sqrt[n]{\lvert c_n \rvert} \to L, and suits coefficients raised to the power n. In practice it is clearest to take the ratio of whole terms, x included, and ask for which x its limit is below 1. That version still works when infinitely many coefficients are 0 (Example 7).

Day 2: the endpoints

At \lvert x - a \rvert = R the ratio limit is exactly 1, the case where the ratio and root tests say nothing. Each endpoint gives a separate numerical series: substitute x = a + R and x = a - R, simplify the terms, and decide with the tests of the unit on series: the nth term test, the integral test and p-series, comparison, and the alternating series test. Two facts organize the work.

• The two endpoints have terms of the same size. At x = a \pm R the terms are c_n (\pm R)^n, of absolute value \lvert c_n \rvert R^n either way. So if one endpoint converges absolutely, so does the other (Exercise 22), and conditional convergence can occur only at an endpoint (Exercise 23). Often one endpoint gives a series of positive terms and the other its alternating version.
• Four shapes occur: (a-R,\, a+R), [a-R,\, a+R], [a-R,\, a+R) and (a-R,\, a+R] (Examples 1, 5 and 6, and Exercise 14).

The same reasoning handles series written in other powers. A series in powers of (bx - c) with b \ne 0 is a power series centered at \displaystyle \frac{c}{b}, since \displaystyle (bx-c)^n = b^n\left(x - \frac{c}{b}\right)^n. The ratio test gives a condition \lvert bx - c \rvert < K, which is \displaystyle \left\lvert x - \frac{c}{b} \right\rvert < \frac{K}{\lvert b \rvert}, so \displaystyle R = \frac{K}{\lvert b \rvert}. A series in powers of (x-a)^2 or (x-a)^3 has zero coefficients in between, so the ratio of consecutive nonzero terms is used, and the condition it gives on (x-a)^2 or (x-a)^3 is turned into one on \lvert x-a \rvert.

Worked examples

Boundaries

• The ratio and root tests are silent at the endpoints. There the limit is exactly 1, and the endpoint series must be decided by other tests. The two endpoints can behave differently (Example 6).
• \displaystyle R = \frac1L needs a limit. The formula needs c_n \ne 0 from some index on and \displaystyle \left\lvert \frac{c_{n+1}}{c_n} \right\rvert to have a limit. With zero coefficients, use the ratio of nonzero terms (Example 7); when no ratio limit exists, the root test may still settle it. The radius itself exists in every case, since the three possibilities hold for every power series.
• The interval is always centered at the center. Convergence spreads inward, so the set where a power series converges is never two separate pieces, and never lopsided apart from its endpoints (Exercise 10).
• A formula for the sum may outlive the series. In Example 1, \displaystyle \frac{5}{3-x} has values outside (-7, 3), but the series does not. The two agree only on the interval of convergence.
• Real x only. This lesson takes x real. When x is allowed to be complex, the same proof with \lvert x - a \rvert read as distance in the complex plane gives convergence inside a disk of radius R, which is where the word radius comes from.

Where this goes next

Inside its interval a power series is a genuine function, and the next lessons treat it as one. Representing functions as power series differentiates and integrates power series term by term, which keeps the radius (Exercise 9 is a first glimpse) though the endpoints can change, and builds series for new functions out of the geometric one. Taylor and Maclaurin series turns the question around: given a function, which power series has it as its sum, with coefficients from derivatives as in Taylor polynomials. The common Maclaurin series finds the series of the common functions, several with R = \infty, and The Taylor remainder and the Lagrange error bound measures how close a partial sum is, which is how one proves that a Taylor series really sums to its function.

Exercises

Day 1: center and radius

For Exercise 1, find the interval of convergence and a formula for the sum. For Exercises 2 to 7, find the center and the radius of convergence, and the open interval on which the series converges absolutely; the endpoints are left for Day 2.

1. \sum_{n=0}^{\infty} \dfrac{(1-x)^n}{4^n}
2. \sum_{n=1}^{\infty} \dfrac{n^3\,(x-2)^n}{5^n}
3. \sum_{n=1}^{\infty} \dfrac{(x+9)^n}{n^n}
4. \sum_{n=1}^{\infty} \dfrac{(n!)^2}{(2n)!}(x-3)^n
5. \sum_{n=1}^{\infty} n^n (x+1)^n
6. \displaystyle \sum_{n=1}^{\infty} \dfrac{(2-3x)^n}{n\sqrt{n}}
7. \sum_{n=1}^{\infty} \left(\dfrac{2n+1}{5n+2}\right)^n (x-1)^n

Day 1: reasoning

8. A power series \sum c_n x^n converges at x = 4 and diverges at x = -6. For each of x = -3, x = 7, x = -4 and x = 5, say whether the series must converge, must diverge, or could do either, and give examples for the last kind.
9. Suppose c_n \ne 0 and \displaystyle \left\lvert \frac{c_{n+1}}{c_n} \right\rvert \to L. Show that \sum c_n (x-a)^n and \sum n\,c_n (x-a)^n have the same radius of convergence.
10. Nima asks whether some power series could converge on (0, 1) and on (2, 3) and nowhere else. Explain why no power series does.

Day 2: intervals of convergence

Find the interval of convergence. At each endpoint, say whether the series converges absolutely, converges conditionally, or diverges.

11. \sum_{n=1}^{\infty} \dfrac{(x-5)^n}{(n+1)\,2^n}
12. \sum_{n=1}^{\infty} \dfrac{(-1)^n (x+8)^n}{n^{3/2}}
13. \sum_{n=1}^{\infty} \dfrac{(4x+1)^n}{n\,2^n}
14. \displaystyle \sum_{n=1}^{\infty} \dfrac{(3-x)^n}{\sqrt{n}}
15. \sum_{n=1}^{\infty} \dfrac{n\,(x-1)^n}{3^n}
16. \sum_{n=1}^{\infty} \dfrac{x^{2n}}{n\,5^n}
17. \sum_{n=0}^{\infty} \dfrac{(-1)^n (x-2)^{2n+1}}{(2n+1)\,4^n}
18. \sum_{n=2}^{\infty} \dfrac{(x-3)^n}{6^n\, n\ln n}
19. \sum_{n=1}^{\infty} \dfrac{8^n (x-1)^{3n}}{n^2}
20. \sum_{n=1}^{\infty} \dfrac{(2n+1)(x-4)^n}{n^3+n}

Day 2: reasoning

21. Kesang wants a power series whose interval of convergence is exactly [1, 7). Give one and justify its interval.
22. Prove that if a power series with radius R, 0 < R < \infty, converges absolutely at one endpoint of its interval, then it converges absolutely at the other.
23. Explain why a power series cannot converge conditionally at any x with \lvert x - a \rvert < R, so that conditional convergence can happen only at an endpoint.