Math Core

A Taylor polynomial has finitely many terms. Let the degree go to infinity and you get a power series: an "infinite polynomial" in $x$ . A power series converges for some values of $x$ and diverges for others, and the set where it converges always has a very particular shape. Finding that set is one of the most dependable free-response questions on the AP exam.

Power series

A power series centered at $x = a$ is a series of the form

\sum_{n=0}^{\infty} c_n (x - a)^n = c_0 + c_1(x - a) + c_2(x - a)^2 + \cdots.

For each particular $x$ , this is an ordinary series of numbers, which either converges or diverges. The geometric series $\sum x^n = \frac{1}{1-x}$ is the simplest example: it converges for $-1 < x < 1$ and diverges everywhere else.

Partial sums of Σ xⁿ (degree 2 and degree 6) hug 1/(1 − x) inside −1 < x < 1 but peel away from it outside that interval.Open in grapher →

The shape of the convergence set

It turns out that every power series behaves like the geometric one in this respect: it converges on an interval centered at $a$ and diverges outside it.

Radius of convergence

For a power series centered at $a$ , exactly one of the following is true:

It converges only at $x = a$ . (Radius $R = 0$ .)
It converges for all real $x$ . (Radius $R = \infty$ .)
There is a number $R > 0$ such that the series converges absolutely for $|x - a| < R$ and diverges for $|x - a| > R$ .

$R$ is the radius of convergence. The interval of convergence is the set of all $x$ where the series converges.

In case 3, the test tells you nothing at the two endpoints $x = a - R$ and $x = a + R$ . The series might converge at both, at one, or at neither, so the interval can look like $(a - R, a + R)$ , $[a - R, a + R)$ , $(a - R, a + R]$ , or $[a - R, a + R]$ .

A power series centered at 1 with R = 2 and interval −1 ≤ x < 3: the radius fixes the interior; the endpoints are checked separately.

Why an interval?

Here is the intuition behind the theorem. If a power series centered at $0$ converges at some point $x = b$ , then its terms $c_n b^n$ go to $0$ , so they are bounded. For any $x$ closer to the center, $|x| < |b|$ , you can write

|c_n x^n| = |c_n b^n|\cdot\left|\frac{x}{b}\right|^n,

which is at most a constant times a geometric term with ratio $\left|\frac{x}{b}\right| < 1$ . So the series converges (absolutely) at every point closer to the center than $b$ . Convergence at one point spreads inward toward the center, and divergence at one point spreads outward. The only possible shape is a symmetric interval around the center, with the two endpoints left undecided.

The radius depends only on how fast the coefficients $c_n$ grow or shrink. Coefficients that shrink very fast, like $\frac{1}{n!}$ , allow convergence everywhere. Coefficients that grow very fast, like $n!$ , allow convergence only at the center. Coefficients like $\frac{1}{2^n}$ give a finite radius, here $R = 2$ .

The procedure

Finding the interval of convergence

Ratio test. Compute $L(x) = \displaystyle\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|$ , where $a_n$ includes the $x$ 's.
Solve $L(x) < 1$ . This gives an inequality $|x - a| < R$ , so you get the radius and the open interval.
Test each endpoint separately by plugging it into the series and using a numerical-series test (p-series, alternating series test, nth term test, comparison).
Assemble the interval, including only the endpoints where the series converges.

Common mistake

The ratio test is always inconclusive at the endpoints (it gives $L = 1$ there), so you must check them another way. Also, when you plug in an endpoint, simplify carefully: $(x - a)^n$ becomes $R^n$ or $(-R)^n = (-1)^n R^n$ , which often cancels a $R^n$ in the denominator and leaves an alternating or p-series.

Worked examples

Worked example: A basic interval

Find the interval of convergence of $\displaystyle\sum_{n=1}^{\infty} \frac{x^n}{n}$ .

Solution. Ratio test:

\left|\frac{x^{n+1}}{n+1}\cdot\frac{n}{x^n}\right| = |x|\cdot\frac{n}{n+1} \to |x|.

Convergence requires $|x| < 1$ , so $R = 1$ .

Endpoints: At $x = 1$ : $\sum \frac{1}{n}$ , the harmonic series, diverges. At $x = -1$ : $\sum \frac{(-1)^n}{n}$ converges by the alternating series test.

Interval: $-1 \le x < 1$ .

Worked example: A shifted center

Find the radius and interval of convergence of $\displaystyle\sum_{n=1}^{\infty} \frac{(x - 3)^n}{n^2\, 2^n}$ .

Solution. Ratio test:

\left|\frac{(x-3)^{n+1}}{(n+1)^2 2^{n+1}}\cdot\frac{n^2 2^n}{(x-3)^n}\right| = \frac{|x - 3|}{2}\cdot\frac{n^2}{(n+1)^2} \to \frac{|x-3|}{2}.

$\frac{|x-3|}{2} < 1$ gives $|x - 3| < 2$ , so $R = 2$ and the open interval is $1 < x < 5$ .

Endpoints: At $x = 5$ : $\sum \frac{2^n}{n^2 2^n} = \sum \frac{1}{n^2}$ converges ( $p = 2$ ). At $x = 1$ : $\sum \frac{(-2)^n}{n^2 2^n} = \sum \frac{(-1)^n}{n^2}$ converges absolutely.

Interval: $1 \le x \le 5$ .

Worked example: Radius infinity and radius zero

Find the radius of convergence of (a) $\displaystyle\sum_{n=0}^{\infty} \frac{x^n}{n!}$ and (b) $\displaystyle\sum_{n=0}^{\infty} n!\,x^n$ .

Solution. (a) $\left|\dfrac{x^{n+1}}{(n+1)!}\cdot\dfrac{n!}{x^n}\right| = \dfrac{|x|}{n+1} \to 0$ for every $x$ . Since $0 < 1$ always, the series converges for all $x$ : $R = \infty$ .

(b) $\left|\dfrac{(n+1)!\,x^{n+1}}{n!\,x^n}\right| = (n+1)|x| \to \infty$ unless $x = 0$ . The series converges only at $x = 0$ : $R = 0$ .

Worked example: A coefficient on x

Find the interval of convergence of $\displaystyle\sum_{n=1}^{\infty} \frac{(2x - 1)^n}{\sqrt{n}}$ .

Solution. Ratio test: $|2x - 1|\cdot\sqrt{\dfrac{n}{n+1}} \to |2x - 1|$ . So we need $|2x - 1| < 1$ , i.e. $0 < x < 1$ . Writing $|2x - 1| = 2\left|x - \frac12\right| < 1$ shows the center is $\frac12$ and $R = \frac12$ .

Endpoints: At $x = 1$ : $\sum \frac{1}{\sqrt n}$ diverges ( $p = \frac12$ ). At $x = 0$ : $\sum \frac{(-1)^n}{\sqrt n}$ converges by the alternating series test.

Interval: $0 \le x < 1$ .

Tip

At an endpoint where the series converges, note whether it converges absolutely or conditionally. AP questions sometimes ask exactly that. In the first example, the series converges conditionally at $x = -1$ .

Practice

Practice 1

Find the interval of convergence of $\displaystyle\sum_{n=1}^{\infty} \frac{x^n}{n}$ .

Enter an inequality, e.g. x >= 4, -2 < x <= 3, or x < 1 or x > 5

Practice 2

Find the radius of convergence of $\displaystyle\sum_{n=1}^{\infty} \frac{(x - 3)^n}{n\,2^n}$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

Find the interval of convergence of $\displaystyle\sum_{n=1}^{\infty} \frac{(x - 3)^n}{n\,2^n}$ .

Enter an inequality, e.g. x >= 4, -2 < x <= 3, or x < 1 or x > 5

Practice 4

Find the radius of convergence of $\displaystyle\sum_{n=0}^{\infty} n!\,(x - 4)^n$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

What is the interval of convergence of $\displaystyle\sum_{n=0}^{\infty} \frac{(x+2)^n}{n!}$ ?

Practice 6

Find the interval of convergence of $\displaystyle\sum_{n=1}^{\infty} \frac{(x + 1)^n}{n^2\, 3^n}$ .

Enter an inequality, e.g. x >= 4, -2 < x <= 3, or x < 1 or x > 5

Practice 7

Find the interval of convergence of $\displaystyle\sum_{n=1}^{\infty} \frac{(2x - 1)^n}{\sqrt{n}}$ .

Enter an inequality, e.g. x >= 4, -2 < x <= 3, or x < 1 or x > 5

Practice 8

Find the interval of convergence of $\displaystyle\sum_{n=1}^{\infty} \frac{n\,x^n}{4^n}$ .

Enter an inequality, e.g. x >= 4, -2 < x <= 3, or x < 1 or x > 5