The interval and radius of convergence of a power series
Bro. David E. Brown, BYU–Idaho Dept. of Mathematics. All rights reserved.
Version 1.12, of April 2, 2014
Contents
1 Introduction
1
2 The ratio and root tests
2
3 Examples
3
4 Exercises
6
5 Answers
7
1
Introduction
P∞
Suppose given a power series n=0 an (x − x0 )n . Recall that the center of the power series is x0 , and that
we say that the series is centered at x0 . Also recall that we say that the series converges if it adds up to a
real number;1 otherwise, we say that the series diverges. Of course, the series might converge at some values
of x and diverge at others. The following definition expresses this more formally.
PN
Definition 1.1. The above series converges at x if and only if limN →∞ n=0 an (x−x0 )n exists and is finite.
At each x for which this limit exists, we make the definition
∞
X
n=0
If, for some x, limN →∞
1∞
PN
n=0
an (x − x0 )n = lim
N →∞
N
X
an (x − x0 )n .
n=0
an (x − x0 )n does not exist, we say that the series diverges at x.
is not a real number.
1
P∞
Every power series n=0 an (x −P
x0 )n converges at its P
center. (To seeP
why, substitute x0 for x in the
∞
∞
∞
n
n
formula of the series, and simplify:
a
(x
−
x
)
=
a
·
0
=
n
0
0
n
n=0
n=0
n=0 0 = 0.) But this leaves us
with the question of whether the series converges at some other x-value. If so, there’s a whole interval of
x-values on which the series converges. This fact is celebrated in the next definition, which is here to remind
you of the meanings of the terms interval of convergence and radius of convergence:
P∞
Definition 1.2. Let n=0 an (x − x0 )n be given.
1. Suppose the series only converges at x = x0 . In this case, the interval of convergence is {x0 }, and the
radius of convergence is R = 0.
2. Suppose series converges at every x. In this case, the interval of convergence is the entire number line,2
and the radius of convergence is R = ∞.
3. Suppose the series converges at each x in some interval (a, b) that is not the entire real line. In this
b−a
.
case, the interval of convergence is (a, b)3 and the radius of convergence is R =
2
This document gives a method for calculating the interval and radius of convergence of a power series.
Sadly, no method can do this for every power series. However, the method given here works well for the
kinds of series found in a typical undergraduate differential equations course.
2
The ratio and root tests
This section gives the promised method for finding the interval of convergence of aP
power series. For reasons
∞
that will become clear later, let An (x) = an (x − x0 )n , so that the power series is n=0 An (x).
The method can make use of any of a number of tools, two of which are now given.
Tool 1: Ratio Test
An+1 (x) , if the limit exists.
Theorem 2.1. Let ρ(x) = lim n→∞
An (x) • The series converges for all values of x for which ρ(x) < 1.
• If ρ(x) = 1, the series may converge or diverge at x.
• The series diverges for all values of x for which ρ(x) > 1.
(Note that ρ is a function of x, because both An+1 (x) and An (x) are; after you take the limit, n no longer
appears, but x still can.4 )
Tool 2: The Root Test
Theorem 2.2. Let ρ(x) = lim
n→∞
q
n An (x), if the limit exists.
• The series converges for all values of x for which ρ(x) < 1.
2 The
number line is known variously as the real line, the interval (−∞, ∞), the real numbers, or (most elegantly), R.
Some authors define the interval of convergence a little differently than this. You see, the series may or may not
converge at x0 − R or at x0 + R, so some people like to insist that you include in your interval of convergence any endpoint at
which the series converges. We don’t need to bother with this. This semester we need power series for differential equations, and
our interest will be in open intervals only. So we’re just using the interior of the interval of convergence, which is (x0 −R, x0 +R),
whether the interval of convergence actually includes any of its endpoints or not.
4 You might be concerned about possibly dividing by 0. If any of the terms in the series are 0, we can throw them out and
rename the rest of the terms. Doing so will not change whether the series converges, and if the series converges, throwing out
the terms that equal zero will not change the sum.
3 Note:
Page 2
• If ρ(x) = 1, the series may converge or diverge at x.
• The series diverges for all values of x for which ρ(x) > 1.
(Again, ρ is a function of x, because An (x) is; after you take the limit, n no longer appears, but x still
can.)
Note that the bulleted items are the same for the root test as for the ratio test.
The Method:
The method consists of the following steps:
1. Calculate ρ(x), whether by using the Root Test or the Ratio Test.
2. Solve the inequality ρ(x) < 1 for x, and you have the interval of convergence.
3. Determine the radius of convergence, based on the interval of convergence:
• If the interval of convergence is {x0 }, the radius of convergence is R = 0.
• If the interval of convergence is the entire real line, the radius of convergence is R = ∞.
• Otherwise, the interval of convergence is the interval (a, b). In this case, the radius of convergence
is r = b−a
2 .
The use of this method is illustrated in the next section.
There are many other tools for finding the interval and radius of convergence of a power series. You can
find some of them in just about any advanced calculus text. Those given above are two of the most basic
ones, but they are versatile enough for our needs this semester.
3
Examples
Example 3.1. Calculate the interval and radius of convergence of
P∞
n=0 (x
− 3)n .
Using the ratio test: The general term of the series is An = (x − 3)n , so An+1 (x) = (x − 3)n+1 . Then
(x − 3)n+1 (x − 3)n (x − 3) An+1 (x) = lim |x − 3| = |x − 3|.
= lim ρ(x) = lim = lim
n→∞
n→∞
An (x) n→∞ (x − 3)n n→∞ (x − 3)n
If we set ρ(x) < 1, we get |x − 3| < 1. Solving this inequality5 yields 2 < x < 4, so the interval of
4−2
convergence is (2, 4). The radius of convergence is R =
= 1.
2
Using the root test: Again, An (x) = (x − 3)n , for all n. So
p
p
p
ρ(x) = lim n |An (x)| = lim n |(x − 3)n | = lim n |x − 3|n = lim |x − 3| = |x − 3|,
n→∞
n→∞
n→∞
n→∞
with the same results as before: You’ll end up solving |x − 3| < 1 and getting (2, 4) for the interval of
convergence and 1 for the radius of convergence.
Comment: In this example, we ended up solving the same inequality when we used the root test as when
we used the ratio test. Sometimes, you may think the ratio test gives you a different inequality to solve
than the root test. Not to worry: When both tests work, they do have to give you the same interval of
convergence (and therefore the same radius of convergence).
5 Don’t remember how to solve such an inequality? Well, recall that |x − 3| is the distance from 0 to x − 3 on the number
line. Therefore, to say |x − 3| < 1 is to say that x − 3 is less that one unit from 0 on the number line. This means that x − 3
has to be between −1 and 1, that is, −1 < x − 3 < 1. Now solve, by adding 3 throughout, to get that 2 < x < 4.
Page 3
Example 3.2. Calculate the interval and radius of convergence of
Using the ratio test: This time, An (x) =
P∞
n=0
x2n
.
n!
x2(n+1)
x2n
, so An+1 (x) =
, and
n!
(n + 1)!
x2(n+1)
n!x2n+2
1 · 2 · 3 · · · · · (n − 1) · n · x2n · x2
x2
An+1 (x)
(n + 1)!
=
=
=
=
.
2n
2n
2n
An (x)
(n + 1)!x
1 · 2 · 3 · · · · · (n − 1) · n · (n + 1) · x
n+1
x
n!
2
x Therefore, ρ(x) = lim
= 0. Um, we’re supposed to solve ρ(x) < 1, which is 0 < 1, which is
n→∞ n + 1
true no matter what the value of x is! In other words, the series converges for all x. That means the
interval of convergence is (−∞, ∞). Naturally, the radius of convergence is R = ∞.
1/n
q
√
x2n x2
n = √
Using the root test:
An (x) = which looks challenging. What could n n! possibly
n
n! n!
turn out to be? Answer: A mess. Don’t use the root test when you have factorials in your series!
P∞
Example 3.3. Calculate the interval and radius of convergence of n=0 (nx)n
(n + 1)n+1 (n + 1)x n+1 = lim x = who-knows-what!? Hmm. . .
Using the ratio test: ρ(x) = lim n→∞ n
n
n→∞
(nx)
n
If you think of (n + 1)n+1 as a polynomial in n, its degree ought to be n + 1, which ought to imply
(n + 1)n+1
→ ∞ as n → ∞. The algebra we’d have to do to verify this is messy and complicated.
nn
This is an example for which the ratio test is not the best tool to use.
q
p
Using the root test: ρ(x) = lim n (nx)n = lim n |nx|n = lim n|x|, which is 0 if x = 0 and ∞
n→∞
n→∞
n→∞
otherwise. So the interval of convergence of this series is {0}, with radius of convergence R = 0.
n
P∞
1
Example 3.4. Calculate the interval and radius of convergence of n=1 1 +
xn
n
n
n
n+1
1
xn =
xn ; after the previous item, the ratio test
Using the ratio test: An (x) = 1 +
n
n
doesn’t seem very attractive here. Let’s try the root test, instead.
v
u
u n + 1 n n+1
n+1
n
Using the root test: t
xn =
|x|, so ρ(x) = lim
|x| = |x|. Solving |x| < 1 tells
n→∞
n
n
n
us that the interval of convergence is (−1, 1), so the radius of convergence must be 1.
n
1
(Comment: It may be of interest to you to know that lim 1 +
= e ≈ 2.71828, the base of the
n→∞
n
natural logarithm function.)
Example 3.5. Calculate the interval and radius of convergence of
P∞
n=0
(3x)n
n!
(3x)n+1
(3x)
An+1 (x)
3x
(n + 1)!
Using the ratio test: An (x) =
, so
=
=
→ 0 as n → ∞, with or without
n
(3x)
n!
An (x)
n+1
n!
the absolute value bars. Once again, there is no restriction on x; the interval of convergence is all of
R, and R = ∞.
n
Page 4
Using the√root test: Actually, don’t use the root test on this one, because doing so would involve dealing
with n n!, which is nontrivial to do.
Example 3.6. This example is a little more advanced and involves somewhat more complicated algebra
than the other examples. You may skip it, the first time you read this document.
P∞
xn (x − 1)n+1
Calculate the interval and radius of convergence of n=1 (−1)n+1
. (Note that this isn’t
n2n
actually a power series. However, the method for finding the interval and radius of convergence still works,
whether you use the root test or the ratio test.)
Using the ratio test: (The algebra is a little hairy, so I’ll put plenty of steps in. but you need to get to
xn (x − 1)n+1
, so
the point where you can skip some of these steps.) An (x) = (−1)n+1
n2n
xn+1 (x − 1)n+2
An+1 (x)
(n + 1)2n+1
=
An (x)
xn (x − 1)n+1
(−1)n+1
n2n
n2n
(−1)n+2 xn+1 (x − 1)n+2
=
n+1
n+1
n
(−1)
(n + 1)2
x (x − 1)n+1
n
xn+1 (x − 1)n+2 2n
·
·
=− n ·
x
(x − 1)n+1 2n+1 n + 1
x · xn (x − 1)(x − 1)n+1 2n
n
=− n ·
·
·
x
(x − 1)n+1
2 · 2n n + 1
1
n
= −x · (x − 1) · ·
2 n+1
n
x(x − 1).
=−
2(n + 1)
(−1)n+2
Therefore,
n
n
1
n
|x(x − 1)|
ρ(x) = lim x(x − 1) = lim
|x(x − 1)| = |x(x − 1)| lim
=
.
n→∞ 2(n + 1)
n→∞ 2(n + 1)
n→∞ n + 1
2
2
|x(x − 1)|
We need to solve the inequality
< 1. Do a little algebra to get −2 < x2 − x < 2 and note
2
that this is really a system of two quadratic inequalities to solve simultaneously:
(
x2 − x + 2 > 0
.
x2 − x − 2 < 0
Let’s solve this system by solving each inequality separately, and taking the x-values that solve both
inequalities.
The usual recommendation for solving a polynomial inequality is based on the following: Think of the
polynomial as being the formula for a function, and consider its graph in the x,y-plane. The zeros of
the polynomial are boundaries between intervals on which the y-values of the polynomial are positive
(above the x-axis) and intervals on which they are negative (below the x-axis). You find the zeros
of your polynomial, and then use “test points” (x-values that are not zeros of the polynomial) to
determine the signs of the polynomials all along the number line.
Let’s follow this course, starting with, say, the second inequality in our system above. The zeros of
x2 − x − 2 are −1 and 2. A convenient test point is x = 0; it’s between the zeros of the polynomial,
and it’s easy to work with. When x = 0, y = −2. Now, we all know what the graph of a parabola
looks like. If the y-values are negative between the two zeros then they are positive everywhere else.
So the solution set of the inequality x2 − 2x − 2 < 0 is (−1, 2).
Page 5
Now for the first inequality in our little system: The zeros of x2 − x + 2 are complex. Complex numbers
do not appear anywhere on the x-axis, so the graph of this polynomial is either always above the x-axis
or always below the x-axis. We note with satisfaction that the y-intercept (which is 02 − 0 + 2 = 2) is
positive, so x2 − x + 2 > 0 for all x; i.e., every real number is a solution of x2 − x + 2 > 0, our first
inequality.
Finally, we need the set of x-values for which both inequalities are true. These are the x’s in the interval
(−1, 2). This interval is the interval of convergence of the given series, and R = 2 − (−1) /2 = 3/2.
v
u
n
n+1 u
|x| |x − 1|(n+1)/n
x
(x
−
1)
n
√
. As n → ∞, this quantity goes to
Using the root test: t(−1)n+1
=
n
n
n2
2
n
x2 − x x(x − 1) x2 − x , which is therefore ρ(x). We need to solve the inequality =
< 1, which
2 2 2
we did for the first half of this example. The rest of this half of this example goes exactly the same
way as in the first half.
P∞
Example 3.7. Calculate the interval and radius of convergence of n=0 (sin n)xn
(sin(n + 1))xn+1 sin(n + 1) Using the ratio test: ρ(x) = lim =
lim
x, which is problematic. You
n→∞ n→∞
(sin n)xn
sin n
see, as n → ∞, there are many times when sin n is close to 0. At these times, sin(n + 1) is not close
to 0, so the fraction sin(n + 1)/ sin n is very large. This prevents the limit from existing, when x 6= 0.
Of course, if x = 0, the sum is 0 and therefore converges. So the interval of convergence contains only
the point 0, and R = 0.
q
p
Using the root test: ρ(x) = lim n (sin n)xn = lim n | sin n||x|. Again, this limit does not exist for
n→∞
n→∞
nonzero x, this time because sin x oscillates forever. The values of | sin n| vary infinitely many times
from just under 1 to just over 0, as n gets larger and larger. So, again, the interval of convergence is
{0} and the radius of convergence is R = 0.
4
Exercises
Calculate the interval and radius of convergence of each of the following series.
P∞
n
Exercise 4.1.
n=0 (2x)
Exercise 4.2.
P∞
x2n+1
(2n + 1)!
Exercise 4.3.
P∞
3n (x + 2)n
n
Exercise 4.4.
P∞
(nx)n
3n
Exercise 4.5.
P∞
(−x)2n
n!
Exercise 4.6.
P∞
n=0
n=0
n=1
n=0
n=0 (−1)
n
x4n+3
(4n + 3)(2n + 1)!
Page 6
5
Answers
If you have questions about these exercises or their answers, please contact me.
Exercise 4.1: Use either the ratio test or the root test. The interval of convergence is (−1/2, 1/2), and
R = 1/2.
Exercise 4.2: Use the ratio test. The interval of convergence is R, and R = ∞.
Exercise 4.3: Use either the ratio test or the root test. The interval of convergence is (5/3, 7/3), and
R = 1/3.
Exercise 4.4: Use the root test. The interval of convergence is {0}, and R = 0.
Exercise 4.5: Use the ratio test. The interval of convergence is R, and R = ∞.
Exercise 4.6: Use the ratio test. The interval of convergence is R, and R = ∞.
Page 7