LECTURE 32: POWER SERIES
MINGFENG ZHAO
March 27, 2015
∞
X
Theorem 1 (Divergence test). If a series
series
∞
X
ak converges, then lim ak = 0. That means, if lim ak 6= 0, then the
k→∞
k=1
k→∞
ak diverges.
k=1
Theorem 2 (Integral test). If f (x) is a continuous, positive, decreasing function for x ≥ 1, let ak = f (k) for k = 1, 2, · · · ,
then
∞
X
Z
ak
∞
and
f (x) dx
1
k=1
either both converge or both diverge. In the case of convergence, the value of the integral is NOT equal to the value of
the series.
Theorem 3 (Ratio Test). Let
∞
X
ak be an infinite series and r = lim
k→∞
k=1
|ak+1 |
, then
|ak |
I. If 0 ≤ r < 1, the series converges.
II. If r > 1, the series diverges.
III. If r = 1, the ration test is inconclusive.
Theorem 4 (Comparison test). Let
∞
X
ak and
k=1
I. If 0 < ak ≤ bk for all k ≥ 1, and
II. If 0 < bk ≤ ak for all k ≥ 1 and
∞
X
bk converges, then
II. If L = 0 and
∞
X
k=1
k=1
ak and
∞
X
k=1
∞
X
k=1
∞
X
k=1
k=1
bk diverges, then
∞
X
ak and
k=1
I. If 0 < L < ∞, then
bk be series with positive terms, then
k=1
Theorem 5 (Limit comparison test). Let
∞
X
∞
X
∞
X
∞
X
ak also converges.
ak diverges as well.
bk be series with positive terms and L = lim
k→∞
k=1
bk either both converge or both diverge.
k=1
bk converges, then
∞
X
ak converges.
k=1
1
bk
, then
ak
2
MINGFENG ZHAO
III. If L = ∞ and
∞
X
bk diverges, then
k=1
∞
X
ak diverges.
k=1
∞
X
Theorem 6 (Absolutely convergence). If the series
|ak | converges, then the series
k=1
∞
X
ak converges.
k=1
Theorem 7 (p-series). Let p be a real number, then

∞
 converges, if p > 1,
X
1
=
 diverges,
kp
if p ≤ 1.
k=1
∞
Theorem 8 (Geometric series). For the geometric sequence {ark }∞
k=0 , then the sequence of partial sums {Sn }n=1 can
be given by:


 na,
if r = 1
Sn =
n
1
−
r

 a·
, if r 6= 1.
1−r
Then



a
, if |r| < 1,
1
−
r
ar =


k=0
diverges,
if |r| ≥ 1.
∞
X
k
Theorem 9 (Telescoping series). For the series
∞
X
(bk − bk+1 ) such that lim bk = 0, then
k→∞
k=1
Theorem 10 (Properties of Convergent Series). a. Suppose
series
∞
X
cak converges and
k=1
b. Suppose
A ± B.
c. Suppose
∞
X
∞
X
cak = c
k=1
ak converges to A and
k=1
∞
X
∞
X
k=1
∞
X
∞
X
k=1
∞
X
ak converges to A and c is a real number, then the
ak = cA.
bk converges to B, then the series
d. If M is a positive integer, then
k=1
ak and
∞
X
(ak ±bk ) converges, and
k=1
bk diverges, then
k=1
∞
X
(bk − bk+1 ) = b1 .
k=1
k=1
k=1
ak converges and
∞
X
∞
X
∞
X
∞
X
(ak ±bk ) =
k=1
(ak ± bk ) diverges.
k=1
ak either both converges or both divergent. In general, whether a
k=M
series converges does not depend on a finite number of terms added to or removed from the series. However, the value
of a convergent series does change if nonzero terms are added or removed.
Example 1. Is the series
∞
X
(k!)2
convergent or divergent?
(2k)!
k=1
LECTURE 32: POWER SERIES
Let ak =
3
(k!)2
, then
(2k)!
ak+1
ak
=
[(k+1)!]2
[2(k+1)]!
(k!)2
(2k)!
By the ratio test, then the series
2
(2k)!
(2k + 2)!
=
(k + 1)!
k!
=
(k + 1)2 ·
=
k 2 + 2k + 1
1
→ ,
4k 2 + 6k + 2
4
·
1
(2k + 2)(2k + 1)
as k → ∞.
∞
X
(k!)2
is convergent.
(2k)!
k=1
∞ h
π n
π n i
X
cos
.
− 2 · − tan
3
6
n=2 √
π 1
π
3
Since cos
= and tan
=
, then
3
2
6
3
Example 2. Evaluate the series
∞ h
X
n=2
cos
π n
3
π n i
− 2 · − tan
6
=
∞ X
n=2
=
=
=
=
=
=
cos
π n
3
−2
∞ X
− tan
n=2
π n
6
√ !n
∞ n
∞
X
X
1
3
−2
−
2
3
n=2
n=2
√ 2
1 2
− 33
2
√
−2·
1 − 12
1 + 33
1
1
√
−2·
2
3+ 3
√
1
3− 3
−2·
2
6
√
3
1
−1+
2
3
√
1
3
− +
2
3
Power series
Definition 1. A power series has the general form
∞
X
ck (x − a)k , where a and ck ’s are real numbers, and x is a variable.
k=0
The ck ’s are the coefficients of the power series, and a is the center of the power series.
4
MINGFENG ZHAO
The set of values of x for which the the series
∞
X
ck (x − a)k converges it its interval of convergence. The radius of
k=0
convergence of the power series, denoted R, is the distance from the center of the series to the boundary point of the
interval of convergence.
Example 3. Look at the power series
∞
X
k=0
∞
X
∞
X
xk , then ck = 1 for all k ≥ 0 and a = 0. For any fixed number x, we know that
k=0
xk is a geometric series, which implies that
∞
X
xk converges if and only if |x| < 1. In the case that |x| < 1, we have
k=0
∞
X
1
. Hence the interval of convergence of the power series
xk is (−1, 1), and the radius of convergence of
x =
1−x
k=0
k=0
∞
X
the power series
xk is R = 1, and
k
k=0
∞
X
k=0
Theorem 11. For the power series
∞
X
∞
X
k=0
xk =
1
,
1−x
for all |x| < 1.
ck (x − a)k , let r = lim
k→∞
|ck+1 |
, then the radius of convergence of the power
|ck |
∞
X
1
series
ck (x − a) is R = . In particular, the power series
ck (x − a)k absolutely converges for all |x − a| < R,
r
k=0
k=0
∞
X
ck (x − a)k diverges for all |x − a| > R.
and the power series
k
k=0
Example 4. Find the radius of convergence of the power series
∞
X
(x − 2)k
√
.
k
k=1
For any fixed x, then
r
√ 1 k+1 k+1
= 1.
r = lim = lim
1
k→∞
k→∞
k
√k ∞
X
(x − 2)k
1
√
Then the radius of convergence of the power series
is R = = 1, which implies that the series converges
r
k
k=1
absolutely for all x such that r = |x−2| < 1 (that is, 1 < x < 3), and the series diverges for all x such that r = |x−2| > 1
(that is, x > 3 or x < 1).
Example 5. Find the radius of convergence of the power series
∞
X
xk
k=1
k!
.
For any fixed x, then
1 (k+1)! k!
1
1 = lim
r = lim
= lim
= 0.
k→∞
k→∞ (k + 1)!
k→∞ k + 1
k!
LECTURE 32: POWER SERIES
Then the radius of convergence of the power series
∞
X
xk
k=1
k!
is R =
5
1
= ∞, which implies that the series converges
0
absolutely for all x.
Example 6. Determine the radius and interval of convergence of the power series
∞
X
k 20 (2x − 4)k
k=0
∞
X
k 20 (2x − 4)k
(2k + 1)!
k=0
=
∞
X
k=0
(2k + 1)!
. Notice that
2k k 20
· (x − 2)k .
(2k + 1)!
So we get
r
=
=
lim
2k+1 (k+1)20
(2(k+1)+1)!
2k k20
(2k+1)!
lim
2k+1 (k + 1)20 (2k + 1)!
·
(2(k + 1) + 1)!
2k k 20
k→∞
k→∞
2k+1 (k + 1)20 (2k + 1)!
·
·
k→∞
2k
k 20
(2k + 3)!
20
1
1
= lim 2 · 1 +
·
k→∞
k
(2k + 3)(2k + 2)
=
=
lim
0
Then the radius of convergence of the power series
of the power series
∞
X
k 20 (2x − 4)k
k=0
(2k + 1)!
∞
X
k 20 (2x − 4)k
k=0
(2k + 1)!
is R =
1
= ∞, and the interval of convergence
0
is (−∞, ∞).
Combining power series
Theorem 12. Suppose the power series
∞
X
k=0
ck xk and
∞
X
dk xk converge to f (x) and g(x), respectively, on an interval
k=0
I, then
I. The power series
∞
X
(ck ± dk )xk converges to f (x) ± g(x) on I.
k=0
II. Let m be an integer such that k + m ≥ 0 for all therms of the power series xm
X
ck x k =
X
ck xk+m . Then this
series converges to xm f (x) for all x 6= 0 in I. When x = 0, the series converges to lim xm f (x).
x→0
III. If h(x) = bxm , where m is a positive integer and b is a nonzero real number, then the power series
∞
X
k=0
converges to the composite function f (h(x)), for all x such that h(x) ∈ I.
ck [h(x)]k
6
MINGFENG ZHAO
Example 7. Find the power series and interval of convergence for
x5
centered at 0.
1−x
In fact, we have
x5
1−x
1
1−x
∞
X
= x5
xk if |x| < 1
= x5 ·
k=0
=
∞
X
xk+5
k=0
=
∞
X
xk .
k=5
So the power series of
∞
X
x5
centered at 0 is
xk and interval of convergence is (−1, 1).
1−x
k=5
Example 8. Find the power series and interval of convergence for
1
centered at 0.
2 + 4x
In fact, we have
1
2 + 4x
=
=
1
2
1
2
1
1 + 2x
1
·
1 − (−2x)
·
∞
=
1X
(−2x)k
2
if | − 2x| < 1, that is, |x| <
k=0
1
2
∞
=
1X
(−2)k xk
2
k=0
=
∞
X
k=0
So the power series of
(−2)k k
x .
2
∞
X
1
(−2)k k
centered at 0 is
x , and interval of convergence is (−1/2, 1/2).
1 + 2x
2
k=0
Example 9. Find the power series and interval of convergence for
In fact, we have
2
1+x
=
=
2
1 + (x − 1) + 1
2
2 + (x − 1)
2
centered at 1.
1+x
LECTURE 32: POWER SERIES
7
1
2
·
2 1 + x−1
2
=
1
1 − − x−1
2
k
∞ X
x − 1
x−1
< 1, that is, |x − 1| < 2
if −
−
2
2 k=0
k
∞ X
1
−
(x − 1)k .
2
=
=
=
k=0
k
∞ X
2
1
(x − 1)k , and interval of convergence is (−1, 3).
centered at 1 is
So the power series of
−
1+x
2
k=0
Differentiating and integrating power series
Theorem 13. Suppose the power series
∞
X
ck (x − a)k converges to f (x) for |x − a| < R, then
k=0
I. Then f (x) is differentiable for |x − a| < R, and
0
f (x) =
∞
X
kck (x − a)k−1 ,
for all |x − a| < R.
k=0
II. Then
Z
f (x) dx =
∞
X
ck
k=0
Example 10. Let f (x) =
(x − a)k+1
+ C,
k+1
for all |x − a| < R.
1
, then
1−x
∞
f (x) =
X
1
xk = 1 + x + x2 + x3 + · · · ,
=
1−x
for all |x| < 1.
k=0
By Theorem 13, we have
Z
Z
f (x) dx
1
dx = − ln |1 − x|
1−x
Z X
∞
=
xk dx
=
k=0
=
∞ Z
X
xk dx
k=0
=
∞
X
xk+1
+C
k+1
k=0
8
MINGFENG ZHAO
∞
X
xk
=
k=1
k
+C
Since − ln |1 − 0| = 0, then C = 0. Hence we have,
ln(1 − x) = −
∞
X
xk
k=1
Remark 1. Recall the series
∞
X
(−1)k
k=1
k
k
,
for all |x| < 1.
conditionally converges, so we get
ln 2 = −
∞
X
(−1)k
k=1
k
=1−
1 1 1 1
+ − + − ··· .
2 3 4 5
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca