23. Power Series
P. K. Lamm
Lecture Notes:
11/11/10
p. 1 / 9
23. Power Series
These classnotes are intended to be supplementary to the textbook and are necessarily limited
by the time allotted for classes. For full and precise statements of definitions and theorems,
as well as material covering other topics and examples, please consult the textbook.
1. Infinite series involving the variable x
Definition: A series of the form
∞
X
cn (x − a)n ,
cn ∈ R, a ∈ R
(1)
n=0
is called a power series about x = a (or just power series).
The special case of a = 0 gives
∞
X
cn x n ,
cn ∈ R,
n=0
a power series about x = 0.
For any value of a, a power series can be viewed as an “infinite polynomial” in the variable x.
If x is assigned a particular value, say, x = x̂, then the power series in (1) reduces to a standard
∞
X
an , where an = cn (x̂ − a)n is just an n-dependent constant for each n = 0, 1, . . ..
infinite series
n=0
It is worth noting that if cn > 0 for all n = 0, 1, . . ., then the following situations hold:
• For specific values of x > a, the power series in (1) is a positive infinite series.
• For x = a, the power series in (1) reduces to the sum c0 + c1 · 0 + c2 · 0 + · · ·, so that the
power series converges and sums to c0 .
• For specific values of x < a, the power series is an alternating series. This is true because
x − a < 0, so that x − a = −|x − a| and
∞
X
n=0
n
cn (x − a) =
∞
X
n
cn (−|x − a|) =
n=0
∞
X
(−1)n cn |x − a|n ,
x < a.
n=0
Example 1.1: Consider the power series
∞
x n
X
(−1)n
.
3
n=0
1
(2)
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For a given value of x ∈ R, this is just a geometric series with r = (−x)/3 . Since a geometric
series converges if |r| < 1 and diverges for |r| ≥ 1, we know that convergence is assured if
−x |x|
3 = 3 < 1,
i.e., if
|x| < 3,
or − 3 < x < 3.
So for all values of x ∈ (−3, 3) the power series in (2) converges, while for |x| ≥ 3 it diverges.
Further, we can write the sum of the power series when x ∈ (−3, 3):
∞
X
(−1)
n
x n
n=0
3
=
3
1
=
,
1 − (−x/3)
3+x
x ∈ (−3, 3).
(3)
The result in (3) is rather remarkable because it says the function
f (x) =
3
,
x+3
x ∈ (−3, 3),
(4)
can be replaced by an infinite series depending on x,
f (x) =
∞
X
(−1)n
x n
3
n=0
,
x ∈ (−3, 3).
Now the equality in (3) is due to the convergence of the power series for x suitably restricted.
∞
X
an converges if the sequence {sn } of partial sums,
Recall that an infinite series
n=0
sn =
n
X
ak ,
k=0
converges, i.e., if sn → L as n → ∞ for some finite constant L. But the nth partial sum sn of a
power series is actually an nth degree polynomial in x which we’ll write as Pn (x). So in this case,
0
P0 (x) = (−1)
x 0
3
1
P1 (x) = 1 − x,
3
1
P2 (x) = 1 − x +
3
..
..
.
.
1
Pn (x) = 1 − x +
3
..
..
.
.
= 1,
1 2
x
9
1 2
1
(−1)n n
x − x3 + · · · +
x .
9
27
3n
2
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So for x ∈ (−3, 3) this means that the rational function f (x) = 3/(x + 3) is approximated by the
nth degree polynomial Pn for n sufficiently large, i.e.,
1
1 2
1 3
(−1)n n
3
≈ 1 − x + x − x + ··· +
x , x ∈ (−3, 3),
x+3
3
9
27
3n
for n sufficiently large. This may be useful information because, compared to rational functions,
polynomials are generally easier to integrate, differentiate, etc..
2. Convergence of power series
All power series share a particular property regarding convergence/divergence, as is detailed in
the following theorem.
Theorem 1 The power series
∞
X
cn (x − a)n behaves in exactly one of the following three ways:
n=0
(i) The series converges for x = a and diverges for x 6= a.
(ii) The series converges absolutely for |x − a| < R and diverges for |x − a| > R, for some value
of R ∈ (0, ∞).
(iii) The series is absolutely convergent for all x ∈ R.
Some comments are in order:
• It is obvious that a power series about a converges absolutely when x = a since in this case
∞
X
n
|cn (x − a) | =
n=0
∞
X
|cn | 0n = |c0 |.
n=0
• One sometimes sees the three possible outcomes from the theorem stated as follows:
The power series converges absolutely for |x − a| < R and diverges for |x − a| > R,
for some R ∈ [0, ∞) ∪ {+∞}.
The case R = 0 corresponds to (i) in the theorem, which the case R = ∞ corresponds to
(iii).
• The constant R is called the radius of convergence for the power series. The interval of
absolute convergence is the set of all x such that |x − a| < R, i.e.,
a − R < x < a + R,
or
x ∈ (a − R , a + R).
• Note that the theorem does not give any information about convergence or divergence of the
power series at the closed endpoints x = a−R and x = a+R of the open interval (a−R, a+R),
and indeed the series may converge or diverge at one or both points. Unfortunately the only
way to establish this is to evaluate the power series at each endpoint, applying an appropriate
test to the infinite series which results. Unless the power series is of a recognizable form (like
a geometric series), we will not take the time to evaluate endpoints because the process for
evaluating convergence of series with constant terms has been considered in detail in recent
lectures.
3
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Example 2.1: Find the interval of absolute convergence for the power series
∞
X
2n (x − 9)n
5n
n=0
,
and find its sum when x is in this interval.
As in the last example, this is a geometric series for each value of x, with the parameters for the
geometric series given by a = 1 and r = 2(x − 9)/5. We therefore have absolute convergence for
|r| < 1, i.e., for
2(x − 9)
< 1,
−1 <
5
or, solving for x,
5
5
− <x−9< ,
2
2
or
5
5
− +9<x<9+ .
2
2
The geometric series diverges for |r| ≥ 1 or for
13
5
x≤− +9=
2
2
x≥9+
and
5
23
= .
2
2
The interval of absolute convergence is thus (13/2, 23, 2), and for any x outside that interval the
power series diverges. Finally, the sum of the power series is given by
∞
X
2n (x − 9)n
n=0
5n
1
5
=
=
,
1 − 2(x − 9)/5
23 − 2x
for x ∈
13 23
,
2 2
.
Example 2.2: Find the (open) interval (x1 , x2 ) of absolute convergence for the power series
∞
X
xn
n=1
n
,
without worrying about convergence/divergence at the endpoints of this interval.
First of all, this power series is not a geometric series so the process is not as direct as in the last
example. We will instead use one of the tests for positive/nonnegative series on the “absolute”
version of this series,
∞
X
|x|n
,
(5)
n
n=1
on the hope that the test will reveal which values of x give convergence. Quite often the ratio test
for positive series will provide all the information that is needed for power series problems (with
the possible exception of behavior at endpoints), so the usual practice is to skip the nth -term test
4
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and go straight to the ratio test. The series in (5) is positive for x 6= 0 so the ratio test leads to
the evaluation of the following x-dependent limit:
ρ = ρ(x) = lim
n→∞
|x|n+1
n+1
1
n
n + 1 = lim |x|
=
lim
|x|
·
= |x|,
·
n
n→∞ |x|n
|x|
n + 1 n→∞
1 + (1/n)
n
(Note that x = 0 corresponds to the infinite series
∞
X
x 6= 0.
0 , which converges absolutely.) The ratio
n=0
test then guarantees convergence of the “absolute” form of the series if |ρ(x)| < 1 or if |x| < 1, and
divergence if |x| > 1. Translating these results to the original series, we have that the original series
converges absolutely for |x| < 1 and diverges for |x| > 1; note that the divergence of the original
series follows from the divergence of the “absolute” series using the arguments in Remark 2.1 from
Lecture Notes #22. So the open interval (x1 , x2 ) of absolute convergence is the interval (−1, 1).
Note also that, although we said that we would not worry about endpoints of intervals of
absolute convergence, in this case convergence/divergence at the endpoints is actually quite easily
determined. Indeed, for the right endpoint x = 1 the series takes the form
∞
X
1
,
n
n=1
a divergent harmonic series (i.e., a p-series with p = 1). For the left endpoint x = −1, the series
is the alternating harmonic series
∞
X
(−1)n
,
n
n=1
which is known from Lecture Notes #22 to converge (but not absolutely).
Example 2.3: Find the (open) interval (x1 , x2 ) of absolute convergence for the power series
∞
X
2n (x + 1)n
n=1
n!
,
(6)
without worrying about convergence/divergence at the endpoints of this interval.
This series is a power series about x = (−1), but it is not in the form of a geometric series. Using
the ratio test on the “absolute” series we evaluate
2n+1 |x + 1|n+1
n!
· n
n→∞
(n + 1)!
2 |x + 1|n
2n+1
n!
|x + 1|n+1
= lim n ·
·
n→∞ 2
(n + 1)! |x + 1|n
2
= lim
|x + 1|
n→∞ n + 1
= 0, x 6= −1,
ρ(x) =
lim
5
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where we exclude the point x = −1 because the series is not positive at that point (required for
the ratio test). However, we already know that the series converges absolutely at x = −1 because
it is the trivial series with all-zero terms in this case.
According to the ratio test in the case of x 6= −1, we have absolute convergence if ρ(x) < 1;
however, we have just seen that ρ(x) = 0 for all nonzero values of x ∈ R, so absolute convergence
is assured for all x ∈ R. Thus,x the radius R of convergence is R = ∞ and the interval of
(absolute) convergence is (−∞, ∞).
Example 2.4: Determine where the following power series converges absolutely:
∞
X
3 n! xn
n=0
As we did for the other examples we apply the ratio test to the “absolute” version
∞
X
3 n! |x|n of
n=0
this series and evaluate
ρ(x) =
lim
n→∞
3(n + 1)! |x|n+1
,
3 n! |x|n
x 6= 0.
Thus
ρ(x) =
lim (n + 1)|x|
n→∞
= ∞,
x 6= 0.
(The excluded point x = 0 is automatically associated with absolute convergence of this series.)
Because ρ(x) is infinite for all nonzero x, we have ρ(x) > 1 for all such x; that is, the original
series diverges when x 6= 0 and it converges (absolutely) when x = 0. The radius R of convergence
is R = 0.
3. Differentiation and integration of power series
We have two theorems on the differentiation and integration of power series.
Theorem 2 Suppose
∞
X
cn (x − a)n converges absolutely on the interval (a − R, a + R) for some
n=0
R > 0. Then for x ∈ (a − R, a + R) the sum defines a function f (x),
f (x) =
∞
X
cn (x − a)n ,
x ∈ (a − R, a + R),
n=0
where f is infinitely-differentiable for x ∈ (a − R, a + R) (i.e., f (k) is defined for k = 0, 1, 2, . . .,
provided x is in this interval). Further,
f
(k)
∞
∞
X
dk X
dk
n
(x) = k
cn (x − a) =
cn k (x − a)n ,
dx n=0
dx
n=0
6
x ∈ (x − R, a + R),
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P. K. Lamm
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where (absolute) convergence of the infinite series on the right-hand side above is obtained at every
point x ∈ (a − R, a + R).
Theorem 3 Suppose
∞
X
cn (x − a)n converges absolutely to a function f (x) on the interval (a −
n=0
R, a + R) for some R > 0. Then for x ∈ (a − R, a + R) the power series
∞
X
n=0
cn
(x − a)n+1
,
n+1
x ∈ (a − R, a + R),
converges absolutely for every x ∈ (a − R, a + R), with sum given by
Z
f (x) dx =
∞
X
n=0
cn
(x − a)n+1
+ C,
n+1
x ∈ (a − R, a + R).
(7)
The power series in (7) converges absolutely for x ∈ (a − R, a + R).
Remark 3.1. There are a couple of important points to observe for the theorems given above:
1. The results in the theorems are only valid for x in the open interval (a−R, a+R) of absolute
convergence. If a given power series happens to converge at one of the endpoints of this open
interval, convergence of the differentiated or integrated power series is not guaranteed at this
point.
2. The theorems are only valid for power series, not for a generic x-dependent infinite series.
For example, the infinite series
∞
X
sin(n! x)
n4
n=1
converges absolutely for all x ∈ (−∞, ∞), as can be easily verified using the direct comparison test. However the same test can be used to show that the series which is formed by
differentiating the above series term by term, i.e.,
∞
X
n! cos(n! x)
n4
n=1
,
diverges for all x ∈ (−∞, ∞). The point is that neither series is a power series.
Example 3.1: (Example 2.1, revisited)
The function
f (x) =
5
,
23 − 2x
7
x 6=
23
2
23. Power Series
P. K. Lamm
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is differentiable and integrable for all x 6= 23/2. But the function is only associated with a power
series on the interval (13/2, 23/2), the (open) interval of absolute convergence for the power series
∞
X 2n (x − 9)n
5
=
,
23 − 2x n=0
5n
x∈
13 23
,
2 2
(ignoring any behavior at endpoints of such intervals).
It follows then that
∞
X n2n (x − 9)n−1
10
=
,
f (x) =
n
(23 − 2x)2
5
n=0
0
and
Z
x∈
13 23
,
2 2
∞
X 2n (x − 9)n+1
5
f (x) dx = − ln(23 − 2x) + C =
+ C,
n
2
(n
+
1)5
n=0
,
x∈
13 23
,
2 2
,
where the convergence is absolute on this interval in each case.
Example 3.2: Consider the power series
∞
X
xn
. Find the (open) interval of convergence of
n!
n=0
this power series and compute its sum for x in this interval.
This series is not of the form of a geometric series, so we once again use the ratio test on the
absolute form of the series, keeping in mind that the ratio test is valid in this case provided x 6= 0:
n!
xn+1
· n
n→∞ (n + 1)! x
xn+1
n!
= lim
·
n
n→∞ x
(n + 1)!
1
= lim x ·
n→∞
n+1
= 0
ρ(x) =
lim
Thus for arbitrary x 6= 0 we have ρ(x) = 0 < 1 so absolute convergence is obtained for all x 6= 0.
In addition, the choice of x = 0 leads to the infinite series
∞
X
xn 1
x
x2
xn
=
+ +
+ ··· +
+ · · ·
= 1,
n!
0!
1!
2!
n!
x=0
n=0
x=0
so that we have convergence in the case of x = 0 as well.
∞
X
xn
, we have absolute convergence for all x ∈ R. We
n!
found the sum of this series in the case of x = 0 above, but how do we find the sum for nonzero
x when the series in question is not a geometric series (our usual way of determining sums)? We
can start by looking at the properties of this sum as spelled out in Theorems 2 and 3.
We conclude that for the infinite series
n=0
8
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Recall that convergence means that there is a finite number f (x) for each x ∈ R for which
∞
X
xn
n=0
= f (x),
n!
and that, from Theorem 2, this function f (x) is infinitely differentiable on the interval (−∞, ∞).
Computing the first derivative of f , we have
!
X
∞
∞
∞
∞
n
X
X
d
x
d xn
nxn−1 X xn−1
0
f (x) =
=
=
=
.
dx n=0 n!
dx n!
n!
(n − 1)!
n=0
n=0
n=1
Note however that
f (x) =
∞
X
xn
n=0
and
n!
=1+
x x2 x3
+
+
+ ···,
1
2!
3!
∞
X
xn−1
x x2 x3
f (x) =
=1+ +
+
+ ···,
(n − 1)!
1
2!
3!
n=1
0
so we have the surprising result that
f 0 (x) = f (x),
x ∈ R.
The only function which behaves in such a manner is
f (x) = Cex ,
x ∈ R,
which comes from applying separation of variables to the differential equation
Z
Z
df
df
= f =⇒
= dx.
dx
f
Finally, we can determine the constant C from the initial condition seen above,
f (0) = 1.
Thus we have C = 1 and
x
f (x) = e =
∞
X
xn
n=0
9
n!
,
all x ∈ R.
(8)