Radius and Interval of Convergence

Part 1 · Hook

Why this matters

Put an

x

into a series, as in

\sum \frac{(x - 1)^n}{n \cdot 4^n}

, and every value of

x

gives a different series of numbers. Some converge and some do not. The values that converge always form one interval around a center, and finding it, endpoints included, is the most tested skill of the unit.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What does the ratio test conclude when its limit is exactly 1?

Nothing: use another test
The series converges
The series diverges
The series converges to 1

Show the answer

A limit of 1 is the case the ratio test leaves open.

Correct: Nothing: use another test:
The series converges:
The series diverges:
The series converges to 1:

2. Does $\sum_{n=1}^{\infty} \frac{(-1)^n}{n}$ converge?

Yes, by the alternating series test
No: its absolute values form the harmonic series
No, by the nth term test
Yes, as a $p$ -series

Show the answer

It alternates and $\frac{1}{n}$ decreases to 0.

Correct: Yes, by the alternating series test:
No: its absolute values form the harmonic series:
No, by the nth term test:
Yes, as a $p$ -series:

3. Solve $|x - 1| < 4$ .

$-3 < x < 5$
$-4 < x < 4$
$-5 < x < 3$
$x < 5$

Show the answer

$-4 < x - 1 < 4$ , so $-3 < x < 5$ .

Correct: $-3 < x < 5$ :
$-4 < x < 4$ :
$-5 < x < 3$ :
$x < 5$ :

Part 4 · See it

See it first

A number line from -5 to 7. The power series sum of (x - 1) to the n over n times 4 to the n is centered at 1 with radius 4, marked by arrows to -3 and 5. A thick bar runs from -3 to 5, with a filled dot at -3, where the series becomes the alternating harmonic series and converges, and an open dot at 5, where it becomes the harmonic series and diverges. Outside the bar the series diverges. The interval of convergence is [-3, 5). — For $\sum \frac{(x - 1)^n}{n \cdot 4^n}$ : center 1, radius 4, and each endpoint tested on its own gives $[-3, 5)$ . LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Apply the ratio test to $\sum a_n(x - a)^n$ , keeping $x$ as a letterthe limit is a multiple of $|x - a|$
Set that limit below 1 and solvethe radius $R$ and the open interval $(a - R, a + R)$ , where the series converges absolutely
At $x = a - R$ and $x = a + R$ the ratio test limit is exactly 1substitute each endpoint, name the series you get, and test it with another test
Each endpoint converges or diverges on its ownwrite the interval with a bracket for an included endpoint and a parenthesis for one left out

Part 6 · Key ideas

Key ideas

A power series $\sum a_n (x - a)^n$ converges only at its center, for every real $x$ , or on an interval of radius $R$ around the center.
The ratio test gives the radius; it never decides the endpoints.
Test each endpoint separately: substitute, simplify, and use the $p$ -series, alternating series or nth term test.
Radius and interval are different answers: $R = 4$ , but the interval is $[-3, 5)$ .

Part 7 · Misconception

A common mistake

The wrong idea: The interval of convergence of $\sum \frac{(x - 1)^n}{n \cdot 4^n}$ is $(-3, 5)$ , from the ratio test.

What actually happens: The ratio test says nothing at $x = -3$ and $x = 5$ . Tested on its own, $x = -3$ gives $\sum \frac{(-1)^n}{n}$ , which converges, so the interval is $[-3, 5)$ .

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

The ratio test on a power series

The ratio test for $\sum_{n=1}^{\infty} \frac{(x + 2)^n}{n^2\,3^n}$
Step	Result
Term	$a_n$ $= \frac{(x + 2)^n}{n^2\,3^n}$
Ratio	$\left\|\frac{a_{n+1}}{a_n}\right\|$ $= \frac{\|x + 2\|}{3} \cdot \frac{n^2}{(n + 1)^2}$
Limit	$\lim_{n \to \infty} \left\|\frac{a_{n+1}}{a_n}\right\|$ $= \frac{\|x + 2\|}{3}$

1. What is the radius of convergence of the series?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

The series converges when $\frac{|x + 2|}{3} < 1$ , that is $|x + 2| < 3$ , so the radius is $R = 3$ .

Answer: 3

2. On which open interval does the ratio test show the series converges?

$(-1, 5)$
$(-3, 3)$
$(-5, 1)$
$\left(-\frac{7}{3}, -\frac{5}{3}\right)$

Show the answer

$|x + 2| < 3$ means $-3 < x + 2 < 3$ , so $-5 < x < 1$ . The endpoints still have to be tested.

$(-1, 5)$ : The center is $x = -2$ , not $2$ : $x + 2 = x - (-2)$ .
$(-3, 3)$ : That is centered at 0; this series is centered at $-2$ .
Correct: $(-5, 1)$ : Right: $|x + 2| < 3$ means $-5 < x < 1$ .
$\left(-\frac{7}{3}, -\frac{5}{3}\right)$ : That uses $\frac{1}{3}$ as the radius; the radius is 3.

3. At the right endpoint, $x = 1$ , what does the series become, and does it converge?

$\sum \frac{1}{n}$ , which diverges as the harmonic series
A series whose ratio test limit is 1, so it converges
$\sum 1$ , which diverges by the nth term test
$\sum \frac{1}{n^2}$ : a convergent $p$ -series

Show the answer

At $x = 1$ , $x + 2 = 3$ and each term is $\frac{3^n}{n^2\,3^n} = \frac{1}{n^2}$ . The $p$ -series with $p = 2 > 1$ converges, so $x = 1$ is included.

$\sum \frac{1}{n}$ , which diverges as the harmonic series: The denominator has $n^2$ , so the terms are $\frac{1}{n^2}$ .
A series whose ratio test limit is 1, so it converges: A ratio test limit of 1 decides nothing; another test is needed.
$\sum 1$ , which diverges by the nth term test: The $3^n$ cancels but the $n^2$ stays: the terms are $\frac{1}{n^2}$ .
Correct: $\sum \frac{1}{n^2}$ : a convergent $p$ -series: Right: $\frac{3^n}{n^2\,3^n} = \frac{1}{n^2}$ .

4. At the left endpoint, $x = -5$ , what happens?

$\sum \frac{(-1)^n}{n^2}$ , which converges absolutely
$\sum \frac{(-1)^n}{n^2}$ diverges, since the signs alternate
$\sum \frac{(-3)^n}{n^2}$ diverges, since its terms grow
$\sum \frac{1}{n^2}$ converges, the same as at $x = 1$

Show the answer

At $x = -5$ , $x + 2 = -3$ and each term is $\frac{(-3)^n}{n^2\,3^n}$ $= \frac{(-1)^n}{n^2}$ . Its absolute values form $\sum \frac{1}{n^2}$ , which converges, so the series converges (absolutely) and $x = -5$ is included.

Correct: $\sum \frac{(-1)^n}{n^2}$ , which converges absolutely: Right: the absolute values form a convergent $p$ -series.
$\sum \frac{(-1)^n}{n^2}$ diverges, since the signs alternate: Alternating signs never cause divergence on their own; here the sizes $\frac{1}{n^2}$ even converge.
$\sum \frac{(-3)^n}{n^2}$ diverges, since its terms grow: The $3^n$ in the denominator cancels: the terms are $\frac{(-1)^n}{n^2}$ .
$\sum \frac{1}{n^2}$ converges, the same as at $x = 1$ : At $x = -5$ the signs alternate: $(x + 2)^n = (-3)^n$ .

5. Give the interval of convergence of $\sum_{n=1}^{\infty} \frac{(x + 2)^n}{n^2\,3^n}$ , with both endpoints tested.

Type the interval, in interval notation such as [1, 5) or as an inequality such as 1 ≤ x < 5.

Show the answer

The ratio test gives $-5 < x < 1$ . At $x = 1$ the series is $\sum \frac{1}{n^2}$ (converges); at $x = -5$ it is $\sum \frac{(-1)^n}{n^2}$ (converges absolutely). The interval of convergence is $[-5, 1]$ .

Answer: $[-5, 1]$

6. Give the interval of convergence of $\sum_{n=0}^{\infty} \frac{x^{2n}}{5^n}$ .

Type the interval, in interval notation such as [1, 5) or as an inequality such as 1 ≤ x < 5.

Show the answer

The series is geometric with ratio $\frac{x^2}{5}$ , so it converges when $\frac{x^2}{5} < 1$ , that is $|x| < \sqrt{5}$ . At $x = \pm\sqrt{5}$ every term is 1, so both endpoints diverge: $\left(-\sqrt{5}, \sqrt{5}\right)$ .

Answer: $(-\sqrt{5}, \sqrt{5})$

Part 9 · Summary

Summary

For a power series centered at

a

, the ratio test gives

|x - a| < R

: the radius

R

and an open interval. The ratio test limit is 1 at both endpoints, so test each one separately and include it only if the series converges there.

Part 10 · Up next

What comes next

Part 11 · Connections