Power Series Week 8

8.6 Power Series Convergence

Definition A power series about c is a function defined by the infinite sum
∞
X
f (x) = an (x − c)n
n=0

where the terms of the sequence {an }∞

n=0 are called the coefficients and c is referred to as the center of
the power series.

Unlike most of the infinite series we’ve previously have seen, Power Series are functions. Therefore we
ask not if they converge, but for which x values do they converge. We see this if we apply the root test for
convergence. Recall that the Root test guarantees absolute convergence if the limit < 1 and divergence if > 1.

r
n
√
lim an (x − c)n = lim n
an · |x − c|
n→∞ n→∞

Notice if x = c then the limit is 0 and the series converges absolutely. So a power series will always converge
for at least one choice of x. The specific coefficients an will determine if that convergence can extend to
other values.

Be Careful The notation can be a bit annoying. We apply the root test to the entire term inside the
n
r which includes the (x − c) . This can be confusing because we previously phrased the root test
summation
n
as lim an but the an carried a different meaning in that context.
n→∞

Theorem / Definition Every power series converges absolutely for the value x = c. For every power series
that converges at more than a single point there is a value R, called the radius of convergence, such that
the series converges absolutely whenever |x − c| < R and diverges when |x − c| > R. A series may converge
or diverge when |x − c| = R. The interval of x values for which a series converges is called the interval of
convergence. By convention we allow R = ∞ to indicate that a power series converges for all x values. If
the series converges only at its center, we will say that R = 0.

Examples
f (n) (c)
(a) Every Taylor series is also a power series. Their centers coincide and an = .
n!
(b) We saw one example of a power series weeks ago. Recall that for |r| < 1, the geometric series converges
and
∞
X a
arn = a + ar + ar2 + ar3 . . . =
n=0
1 − r

We can view this as a power series where {an } = a and c = 0,

∞
X a
axn =
n=0
1−x

This converges for |x − 0| = |x| < 1 so the radius of converges is 1. The series does not converge for
x = 1 or x = −1. Therefore the interval of convergence is (−1, 1).

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 Power Series Week 8

Testing a Power Series for Convergence

Step 1. Use the ratio test or root test to determine the radius of convergence.

|x − c| < R

Step 2. Plug in x = c + R and x = c − R. Check if the series converges for either of these end points.
Step 3. Write out the interval of convergence ([c−R, c+R)] with the appropriate end point signifier chosen.
Examples Determine the interval of convergence.
∞
X xn
(a) √ n
n=1
n n3

By the ratio test the series will converge when

√ √
an+1 xn+1 n n3n n n 1
lim | | < 1 =⇒ lim √ n+1
· n
< 1 =⇒ lim √ x <1
n→∞ an n→∞ (n + 1) n + 13 x n→∞ (n + 1) n + 1 3

x
=⇒ | | < 1 =⇒ |x| < 3
3

Therefore the radius of convergence is 3 . We now check the end points. When x = 3 we have the series

∞
X 1
√
n=1
n n

which converges as a p series with p > 1. When x = −3 we have the series,

∞
X 1
(−1)n √
n=1
n n

which converges by the alternating series test. This gives that Interval of Convergence: [-3,3] .
∞
X 2n n
(b) (4x − 8)
n=1
n
This strictly isn’t expressed as a power series. But we can factor out a 4 from the parenthetical statement
to give a series centered at 2,
∞
X 8n n
(x − 2)
n=1
n
√
n
Remember that lim n = 1. Let’s try the root test on this. The series will converge absolutely when
n→∞
r
p n 8n n 8 1
lim n |an | = lim (x − 2) = lim √ (x − 2) = |8(x − 2)| < 1 =⇒ |x − 2| <
n→∞ n→∞ n n→∞ n
n 8
1
=⇒ Radius of Convergence:
8

Now check the endpoints. The left and right endpoints respectively give the series,

∞ ∞
X 1n
X 1
(−1)
n=1
n n=1
n

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 Power Series Week 8

The former is the alternating harmonic series and converges. The latter is the harmonic series and
diverges.
 
1 1 15 17
Interval of Convergence: [2 − , 2 + ) = ,
8 8 8 8

∞
X 2
(c) 2n (x − 3)n
n=0

Apply the ratio test,

2
2n +2n+1
(x − 3)n+1
lim = lim 22 n + 1(x − 3) = |x − 3| lim 22n+1
n→∞ 2n2 (x − 3)n n→∞ n→∞

This limit goes to infinity of course. Unless, the number we are multiplying by is zero. Therefore this
converges only when x = 3. Therefore

Radius of Convergence: 0 Interval of Convergence: [3, 3]

∞
X (x + 1)n
(d)
n=0
n!

Note that the center of this power series is −1. Use the ratio test

(x + 1)n+1 n! x+1
lim · = lim =0<1
n→∞ (n + 1)! (x + 1)n n→∞ n + 1

This series converges no matter what the value of n is. Therefore,

Radius of Convergence: ∞ Interval of Convergence: (−∞, ∞)

∞
X
(e) (−1)n nxn
n=0

The center of the power series is 0. Apply the ratio test

(−1)n+1 (n + 1)xn+1 n+1

lim = lim · x = |x| < 1
n→∞ (−1)n nxn n→∞ n

We’ve found that the radius of convergence is 1. We must not Px = ±1

P plug in to determine the behavior
at the end points. We can see that both the resulting series, n & (−1)n n, diverge by the divergence
test. Therefore,

Radius of Convergence: 1 Interval of Convergence: (−1, 1)

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 Power Series Week 8

8.7 New from Old: Representing Functions by Power Series

Now that we know a bit about power series how can we find the power series for a particular function? So
1 P n
far we’ve only seen one really, = x for |x| < 1. We want to milk this identity for as much as we
1−x
can. One way we can do this is by arithmetic operations and substitution.

Examples Determine the power series representation for the given function. Give the interval of convergence
for the series.
x2
(a)
1−x

We don’t have much to work with right now. Let’s stare at the power series we know and see if there’s
a way to transform it into the one we want. Notice that we can do this by multiplication.

∞ ∞
x2 1 X X
= x2 · = x2 xn = xn+2
1−x 1−x n=0 n=0

There we have it. Our power series representation. Let’s think about the interval of convergence. The
base series had the interval (−1, 1). We multiplied that series by a value not depending on n. Hence the
convergence of the series is not affected. Our interval of convergence is then (−1, 1).
1
(b)
1 + 5y
1
Notice we can get this function by plugging in x = −5y into . Therefore,
1−x
∞ ∞
1 1 X X
= = = (−5y)n = (−1)5n y n
1 + 5y 1 − (−5y) n=0 n=0

The original series converges when |x| < 1 so after our substitution our series will converge for | − 5y| <
1 =⇒ |y| < 51 . There’s nothing special about having chosen y for our variable so we could instead say,
∞  
1 X 1
= (−1)n 5n xn for |x| <
1 + 5x n=0 5

1
(c)
2−z

Notice that,
∞ ∞
1 1 1 1 X  z n X 1 n
= · = · = z
2−z 2 1 − ( z2 ) 2 n=0 2 n=0
2n+1

This will converge when

z
= |x| < 1 =⇒ |z| < 2
2
Note 1: When we substitute into a power series we must ensure that the range of our substituted function
falls properly within the interval of convergence of the series.

Note 2: At this point we should take care to only perform substitutions whose result is also a power series.
Meaning, for example, we would not make a substitution such as x = eu or x = sin θ.

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 Power Series Week 8

8.8 Term by Term Differentiation

∞
X
Theorem Assume the power series an (x − c)n converges for |x − c| < R for some R > 0. For x values
n=0
in this interval we can define a function
∞
X
f (x) = an (x − c)n
n=0

f (x) has derivatives of all orders and they are given by differentiating the sum term by term. That is
∞
X
f 0 (x) = nan (x − c)n−1
n=0
Furthermore, the derivative’s power series has the same radius of convergence as the original.

Note The ability to differentiate term by term is a special property of power series. For example,
∞
X sin(n!x)
f (x) =
n=1
n2
1
The series defining this function converges for all values of x (compare it to n2 ). If we take the derivative of
each term we get the series,
∞
X cos(n!x)
n!
n=1
n2
which diverges for all x by the divergence test.

Example Discover the derivative for ex .

Start with the power series

∞
x x2 x3 X xn
e =1+x+ + + ... =
2! 3! n=0
n!
and take the derivative
∞ ∞ ∞ ∞
d x x2 X xn X xn−1 X xn−1 X xn
[e ] = 0 + 1 + x + + ... = n· = n = =
dx 2! n=0
n! n=1
n! n=1
(n − 1)! n=0 n!
Differentiating the power series resulted in the same series which we started with, the series for ex . This is
d x
telling us then that [e ] = ex .
dx

Examples Determine a power series for the given function.

1
(a)
(1 − x)2

Let’s use our shiny new toy in the form of the theorem of this section. We want to think if we know a
function whose derivative will give us the above function. We do.
  ∞ ∞ ∞
1 d 1 d X n X n−1 X
= = x = nx = (n + 1)nxn (for − 1 < x < 1)
(1 − x)2 dx 1 − x dx n=0 n=0 n=0

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 Power Series Week 8

k!
(b)
(1 − x)k+1

Notice that
∞ ∞ ∞
dk dk X n X
 
k! 1 n! X (n + k)! n
k+1
= k = k x = xn−k = x (for − 1 < x < 1)
(1 − x) dx 1−x dx n=0 (n − k)! n=0
n!
n=k

(c) cos x.
∞
X x2n+1
We saw that the power series for sin x is given by (−1)n for −∞ < x < ∞. We can obtain
n=0
(2n + 1)!
the power series expansion for cos x by differentiating.

∞
x3 x5 x7 X x2n+1
sin x = x − + − + ... = (−1)n
3! 5! 7! n=0
(2n + 1)!

∞
x2 x4 x6 X x2n
=⇒ cos x = 1 − + − + ... = (−1)n
2! 4! 6! n=0
(2n)!

The expansion will be valid for −∞ < x < ∞.

Note The theorem says that the radius of convergence is unaffected by differentiation. However, the end
points of convergence may be affected.

∞
X xn
f (x) =
n=1
n2
converges for −1 ≤ x ≤ 1. It’s radius of convergence is 1 and it converges at both endpoints. On the other
hand,
∞
X xn−1
f 0 (x) =
n=1
n
has radius of converges 1 but converges for −1 ≤ x < 1. Since setting x = 1 gives us the harmonic series
which diverges.

8.9 Term by Term Integration

∞
X
Theorem Assume the power series an (x − c)n converges for |x − c| < R for some R > 0. For x values
n=0
in this interval we can define a function
∞
X
f (x) = an (x − c)n
n=0

f (x) is integrable and

Z ∞
X an
f (x) dx = C + (x − c)n+1
n=0
n + 1
Furthermore, the integral’s power series has the same radius of convergence as the original.

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 Power Series Week 8

Note The ability to integrate term by term is a special property of power series. For example,
∞ x
X sin( n! )
f (x) = 2
n=1
n
1
The series defining this function converges for all values of x (compare it to n2 ). If we take the derivative of
each term we get the series,
∞ x
X cos( n! )
− n! 2
n=1
n
which diverges for all x by the divergence test.

Examples Determine a power series for the given function.

(a) arctan x

It would be hard to has this series come as the result of a substitution or differentiation. However we
1
can recognize it as the integral of . And
1 − x2
∞
1 X
= (−1)n x2n (for |x| < 1)
1 − (−x2 ) n=0

by substitution into the geometric series. Now,

∞ ∞
x2n+1
Z Z X
1 X
arctan x = dx = (−1)n x2n dx = C + (−1)n
1 − (−x2 ) n=0 n=0
2n + 1

We can determine C by evaluating the arctan function. arctan 0 = 0 so we can conclude that C = 0.
Therefore,
∞
X x2n+1
arctan x = (−1)n (for |x| ≤ 1)
n=0
2n + 1

Note The theorem guarantees that this series converges for −1 < x < 1. However, we can see that in
fact this extends to endpoints. The behavior at end points can change. Only the radius of convergence
is persistent.
(b) ln(1 + x)

First see that

∞ ∞
1 1 X X
= = (−x)n = (−1)n xn (for |x| < 1)
1+x 1 − (−x) n=0 n=0

and so

∞ Z ∞
X X (−1)n n+1
ln(1 + x) = (−1)n xn dx = C + x (for |x| < 1)
n=0 n=0
n+1

Since ln(1 + 0) = 0 we can conclude that C = 0. Checking end points we see that x = −1 gives the
harmonic series which diverges. x = 1 gives the alternating harmonic series which converges. Putting
this together,

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 Power Series Week 8

∞
X (−1)n n+1
ln(1 + x) = x (for − 1 ≤ x < 1)
n=0
n

2
ex dx.
R
Example Find

Your Math 122 methods of integration will not perform so well on this function. However now that we
2
have power series this is trivial. To start we have to find the power series expansion of ex . Recall that
∞
X xn
2 3
ex = 1 + x + x2! + x3! + . . . = and so
n=0
n!
∞ ∞
2 x4 x6 X (x2 )n X x2n
ex = 1 + x2 + + + ... = =
2! 3! n=0
n! n=0
n!
Integrating this gives our solution,

∞
x3 x5 x7 x2n+1
Z
2 X
ex dx = x + + + + ... =
3 5 · 2! 7 · 3! n=0
(2n + 1) · n!

We can’t recognize this power series as the representation of any function that we’ve encountered before but
that doesn’t make this any less valid of an answer.