Radius and Interval of Convergence

A power series is like a polynomial that never ends: $c_0 + c_1(x - a) + c_2(x - a)^2 + \cdots$. For each value of $x$ it becomes an ordinary series of numbers, which might converge or diverge. The set of $x$-values where it converges is always an interval centred at $a$, and finding that interval is a standard AP BC free-response task. It’s also what tells you where a Taylor series can actually be trusted.

Key ideas

Power series

A power series centred at $x = a$ (or “about $x = a$”) has the form

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

The $c_n$ are constants (the coefficients). At $x = a$, every term after the first is $0$, so every power series converges at its centre.

Only three possibilities

For any power series, exactly one of these is true:

1. It converges only at $x = a$. The radius of convergence is $R = 0$.
2. It converges for all real $x$. The radius is $R = \infty$.
3. There’s a number $R \gt 0$ (the radius of convergence) such that the series converges absolutely when $|x - a| \lt R$ and diverges when $|x - a| \gt R$. At the two endpoints $x = a - R$ and $x = a + R$, anything can happen.

The interval of convergence of $\sum \dfrac{(x - 2)^n}{n \cdot 3^n}$ is $-1 \le x \lt 5$: centre $2$, radius $3$, one endpoint in and one out.

Finding the radius with the ratio test

Apply the ratio test to the absolute value of the whole term, $x$ included:

$$L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|$$

The limit usually comes out as (a number) $\times\, |x - a|$. The series converges absolutely when $L \lt 1$; solve that inequality for $|x - a|$ to read off $R$. If $L = 0$ for every $x$, then $R = \infty$. If $L = \infty$ for every $x \ne a$, then $R = 0$.

Checking the endpoints

At the endpoints the ratio test gives $L = 1$, which tells you nothing. So substitute each endpoint into the series and test the resulting series of numbers with another test:

• $p$-series or a comparison test for positive terms,
• the alternating series test when the signs alternate,
• the $n$th term test when the terms don’t go to $0$.

Writing the interval

The interval of convergence is the set of all $x$ where the series converges, endpoints included or not. Write it as an inequality, such as $-1 \le x \lt 5$ (some books write $[-1, 5)$). The radius is a single number, $R = 3$; the interval is a set of $x$-values. Don’t mix them up.

On the AP exam, “Find the interval of convergence. Justify your answer.” means: show the ratio test, state the radius, and name the test you used at each endpoint.

Worked examples

Example 1: Converges everywhere

Find the interval of convergence of $\displaystyle\sum_{n=0}^{\infty} \frac{x^n}{n!}$.

Solution. Apply the ratio test:

$$\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{x^{n+1}}{(n+1)!} \cdot \frac{n!}{x^n}\right| = \frac{|x|}{n + 1}$$ $$L = \lim_{n \to \infty} \frac{|x|}{n + 1} = 0 \lt 1 \quad \text{for every } x$$

The series converges for all real $x$, so $R = \infty$. (This is the Maclaurin series for $e^x$.)

Example 2: One endpoint in, one out

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

Solution. Ratio test:

$$\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{(x - 2)^{n+1}}{(n+1)3^{n+1}} \cdot \frac{n \cdot 3^n}{(x - 2)^n}\right| = \frac{n}{n + 1} \cdot \frac{|x - 2|}{3}$$ $$L = \lim_{n \to \infty} \frac{n}{n + 1} \cdot \frac{|x - 2|}{3} = \frac{|x - 2|}{3}$$

Converges absolutely when $\dfrac{|x - 2|}{3} \lt 1$, that is, $|x - 2| \lt 3$. So the centre is $2$, $R = 3$, and the open interval is $-1 \lt x \lt 5$.

Endpoint $x = 5$: the series is $\displaystyle\sum \frac{3^n}{n \cdot 3^n} = \sum \frac{1}{n}$, the harmonic series, which diverges.

Endpoint $x = -1$: the series is $\displaystyle\sum \frac{(-3)^n}{n \cdot 3^n} = \sum \frac{(-1)^n}{n}$. This alternates, and $\dfrac{1}{n}$ decreases to $0$, so it converges by the alternating series test.

The interval of convergence is $-1 \le x \lt 5$.

Example 3: A coefficient on x

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

Solution.

$$L = \lim_{n \to \infty} \left|\frac{(2x - 1)^{n+1}}{\sqrt{n + 1}} \cdot \frac{\sqrt{n}}{(2x - 1)^n}\right| = \lim_{n \to \infty} \sqrt{\frac{n}{n + 1}}\,|2x - 1| = |2x - 1|$$

Converges when $|2x - 1| \lt 1$, so $-1 \lt 2x - 1 \lt 1$, which gives $0 \lt x \lt 1$. Writing $|2x - 1| = 2\left|x - \tfrac{1}{2}\right| \lt 1$ shows the centre is $\tfrac{1}{2}$ and $R = \tfrac{1}{2}$ (not $1$).

Endpoint $x = 1$: $\displaystyle\sum \frac{1}{\sqrt{n}}$ is a $p$-series with $p = \tfrac{1}{2} \le 1$, so it diverges.

Endpoint $x = 0$: $\displaystyle\sum \frac{(-1)^n}{\sqrt{n}}$ alternates, and $\dfrac{1}{\sqrt{n}}$ decreases to $0$, so it converges.

The interval of convergence is $0 \le x \lt 1$.

Example 4: Only even powers, both endpoints in

Find the interval of convergence of $\displaystyle\sum_{n=1}^{\infty} \frac{x^{2n}}{4^n n^2}$.

Solution. Keep the whole term, including $x^{2n}$:

$$L = \lim_{n \to \infty} \left|\frac{x^{2n+2}}{4^{n+1}(n+1)^2} \cdot \frac{4^n n^2}{x^{2n}}\right| = \lim_{n \to \infty} \frac{n^2}{(n+1)^2} \cdot \frac{x^2}{4} = \frac{x^2}{4}$$

Converges when $\dfrac{x^2}{4} \lt 1$, so $|x| \lt 2$: centre $0$, $R = 2$.

Endpoints $x = \pm 2$: $x^{2n} = 4^n$ either way, so the series is $\displaystyle\sum \frac{1}{n^2}$, a $p$-series with $p = 2 \gt 1$, which converges.

The interval of convergence is $-2 \le x \le 2$.

Common mistakes

Not checking the endpoints. The ratio test only gives the open interval. Each endpoint must be tested separately, and the answer can be different at each one, as in Example 2.

Using the ratio test at an endpoint. At an endpoint, the ratio test limit is always $1$: inconclusive. Use a $p$-series, comparison, alternating series, or $n$th term test.

Dropping the absolute value. The condition is $|x - a| \lt R$, which gives two inequalities, $a - R \lt x \lt a + R$. Without absolute values you’d only find half the interval.

Getting the radius wrong when x has a coefficient. From $|2x - 1| \lt 1$ the radius is $\tfrac{1}{2}$, because $|2x - 1| = 2\left|x - \tfrac{1}{2}\right|$. Solve for $x$ to find the interval, then read off the centre and radius from it.

Leaving x out of the ratio. In Example 4, the $x^{2n}$ has to be in the ratio, giving $\dfrac{x^2}{4}$. Using only the coefficients would give the wrong radius.

Mixing up the radius and the interval. $R$ is a distance (one number). The interval is a set of $x$-values, like $-1 \le x \lt 5$.

Practice

1. (Warm-up) For a power series, the ratio test gives $L = \dfrac{|x - 3|}{4}$. What are the centre and radius of convergence, and on what open interval does the series converge absolutely?

Solution

$\dfrac{|x - 3|}{4} \lt 1$ gives $|x - 3| \lt 4$. The centre is $3$, $R = 4$, and the series converges absolutely for $-1 \lt x \lt 7$. (The endpoints still need checking.)

2. (Warm-up) Find the interval of convergence of $\displaystyle\sum_{n=0}^{\infty} \frac{x^n}{2^n}$.

Solution

This is geometric with ratio $\dfrac{x}{2}$, so it converges exactly when $\left|\dfrac{x}{2}\right| \lt 1$, that is, $|x| \lt 2$. (The ratio test gives the same: $L = \dfrac{|x|}{2}$.)

At $x = 2$ the series is $\sum 1$, and at $x = -2$ it is $\sum (-1)^n$. In both cases the terms don’t approach $0$, so both diverge by the $n$th term test.

The interval of convergence is $-2 \lt x \lt 2$.

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

Solution

$$L = \lim_{n \to \infty} \frac{n}{n + 1}\,|x - 1| = |x - 1|$$

So $|x - 1| \lt 1$, giving $0 \lt x \lt 2$.

At $x = 2$: $\sum \dfrac{1}{n}$, the harmonic series, diverges. At $x = 0$: $\sum \dfrac{(-1)^n}{n}$ converges by the alternating series test.

The interval of convergence is $0 \le x \lt 2$.

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

Solution

$$L = \lim_{n \to \infty} \left|\frac{(n + 1)(x + 2)^{n+1}}{3^{n+1}} \cdot \frac{3^n}{n(x + 2)^n}\right| = \lim_{n \to \infty} \frac{n + 1}{n} \cdot \frac{|x + 2|}{3} = \frac{|x + 2|}{3}$$

So $|x + 2| \lt 3$, giving $-5 \lt x \lt 1$.

At $x = 1$: $\sum n$, and at $x = -5$: $\sum (-1)^n n$. In both, the terms don’t approach $0$, so both diverge by the $n$th term test.

The interval of convergence is $-5 \lt x \lt 1$.

5. (Core) Find the interval of convergence of $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n x^n}{n^2 \cdot 5^n}$.

Solution

$$L = \lim_{n \to \infty} \frac{n^2}{(n + 1)^2} \cdot \frac{|x|}{5} = \frac{|x|}{5}$$

So $|x| \lt 5$.

At $x = 5$: $\sum \dfrac{(-1)^n}{n^2}$. At $x = -5$: $\sum \dfrac{(-1)^n (-1)^n}{n^2} = \sum \dfrac{1}{n^2}$. In both cases the series of absolute values is $\sum \dfrac{1}{n^2}$, a convergent $p$-series ($p = 2$), so both converge (absolutely).

The interval of convergence is $-5 \le x \le 5$.

6. (Core) Find the radius of convergence of $\displaystyle\sum_{n=0}^{\infty} n!\,(x - 1)^n$.

Solution

$$\left|\frac{a_{n+1}}{a_n}\right| = \frac{(n + 1)!\,|x - 1|^{n+1}}{n!\,|x - 1|^n} = (n + 1)|x - 1|$$

For any $x \ne 1$, this goes to $\infty$ as $n \to \infty$, so the series diverges. It converges only at its centre, $x = 1$. The radius is $R = 0$.

7. (Core) Find the centre, radius and interval of convergence of $\displaystyle\sum_{n=0}^{\infty} \frac{(2x + 3)^n}{n + 1}$.

Solution

$$L = \lim_{n \to \infty} \frac{n + 1}{n + 2}\,|2x + 3| = |2x + 3|$$

$|2x + 3| \lt 1$ gives $-1 \lt 2x + 3 \lt 1$, so $-2 \lt x \lt -1$. The centre is $-\tfrac{3}{2}$ and $R = \tfrac{1}{2}$.

At $x = -1$: $\sum \dfrac{1}{n + 1} = 1 + \dfrac{1}{2} + \dfrac{1}{3} + \cdots$, the harmonic series, diverges.

At $x = -2$: $\sum \dfrac{(-1)^n}{n + 1}$ converges by the alternating series test.

The interval of convergence is $-2 \le x \lt -1$.

8. (Challenge) Find the interval of convergence of $\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n + 1}$.

Solution

$$L = \lim_{n \to \infty} \left|\frac{x^{2n+3}}{2n + 3} \cdot \frac{2n + 1}{x^{2n+1}}\right| = \lim_{n \to \infty} \frac{2n + 1}{2n + 3}\,x^2 = x^2$$

So $x^2 \lt 1$, that is, $|x| \lt 1$.

At $x = 1$: $\sum \dfrac{(-1)^n}{2n + 1} = 1 - \dfrac{1}{3} + \dfrac{1}{5} - \cdots$ converges by the alternating series test.

At $x = -1$: $(-1)^{2n+1} = -1$, so the series is $\sum \dfrac{(-1)^{n+1}}{2n + 1}$, which also converges by the alternating series test.

The interval of convergence is $-1 \le x \le 1$. (This is the series for $\arctan x$; see representing functions as power series.)

9. (Challenge) A power series $\displaystyle\sum_{n=0}^{\infty} c_n (x - 2)^n$ converges at $x = 6$ and diverges at $x = -3$. For each value, decide whether the series must converge, must diverge, or whether you can’t tell: (a) $x = 0$, (b) $x = 8$, (c) $x = -2$.

Solution

Converging at $x = 6$, which is $4$ from the centre, means $R \ge 4$. Diverging at $x = -3$, which is $5$ from the centre, means $R \le 5$. So $4 \le R \le 5$.

(a) $x = 0$ is $2$ from the centre, and $2 \lt 4 \le R$, so the series must converge (absolutely).

(b) $x = 8$ is $6$ from the centre, and $6 \gt 5 \ge R$, so the series must diverge.

(c) $x = -2$ is $4$ from the centre. If $R = 4$, it is an endpoint, and an endpoint can go either way, even though the other endpoint $x = 6$ converges. So you can’t tell.