Lecture 24Section 11.4 Absolute and Conditional
Convergence; Alternating Series
Jiwen He
1
Convergence Tests
Basic Series that Converge or Diverge
Basic Series that Converge
X
Geometric series:
xk ,
X 1
,
kp
p-series:
if |x| < 1
if p > 1
Basic Series that Diverge
Any series
X
p-series:
ak
for which lim ak 6= 0
k→∞
X 1
,
kp
if p ≤ 1
Convergence Tests (1)
Basic Test for Convergence
P
Keep in Mind that,
if ak 9 0, then the series
ak diverges;
therefore there is no reason to apply any special convergence test.
P k
P
Examples 1.
x with |x| ≥ 1 (e.g,
(−1)k ) diverge since xk 9 0. [1ex]
k
X
X k
1
k
diverges since k+1 → 1 6= 0. [1ex]
1−
diverges since
k+1
k
k
ak = 1 − k1 → e−1 6= 0.
Convergence Tests (2)
Comparison Tests
Rational terms
easily handled by basic comparison or limit comparison
Pare most
with p-series
1/k p
Basic Comparison Test
1
X 1 X
k3
1
converges by comparison with
converges
3
5
+1
k
k + 4k 4 + 7
X 1 X
X 2
1
by comparison with
converges by comparison with
2
3
2
k
k −k
k3
X 1
X
X
1
1
diverges by comparison with
diverges by
3k + 1
3(k + 1)
ln(k + 6)
X 1
comparison with
k+6
X
2k 3
Limit Comparison Test
X 1
X 1
X 3k 2 + 2k + 1
converges by comparison with
.
diverges
3
3
k −1
k
k3 + 1
√
X 3 X 5 k + 100
√
√ converges by comparison with
by comparison with
k
2k 2 k − 9 k
X 5
2k 2
Convergence Tests (3)
Root Test and Ratio Test
The root test is used only if powers are involved.
Root Test
X k2
X 1
2
1/k
1/k
= 21 · k 1/k → 12 · 1
=
converges: (ak )
converges: (ak )
k
2
(ln k)k
2
k
k
X
1
1/k
(−1)
1
→
0
converges:
(a
)
=
1
+
→ e−1
1
−
k
ln k
k
k
Convergence Tests (4)
Root Test and Ratio Test
The ratio test is effective with factorials and with combinations of powers and
factorials.
Ratio Comparison Test
X k2
X 1
2
1
converges: aak+1
converges: aak+1
= 21 · (k+1)
→ 12
= k+1
→0
2
k
k
k
k
2
k!
k
X
X k
k
k
1 k+1
1
= 10
= 1 + k1 → e
converges: aak+1
· k → 10
diverges: aak+1
k
k
10k
k!
X 2k
X 1
1−(2/3)k
1
√ converges: aak+1
converges: aak+1
=
2
·
=
k → 2 · 3
3−2(2/3)
k
k
k
k
3
−
2
k!
q
1
k+1
2
2.1
→0
Absolute Convergence
Absolute Convergence
Absolute Convergence
2
Absolute
P Convergence
P
A series
ak is said to converge absolutely if
|ak | converges.
P
P
|ak | converges, then
ak converges.
if
i.e., absolutely convergent series are convergent.
Alternating p-Series with p > 1
X (−1)k
X 1
,
p
>
1,
converge
absolutely
because
converges.
kp
kp
∞
k+1
X (−1)
1
1
1
= 1 − 2 + 2 − 2 − · · · converge absolutely.
2
k
2
3
4
⇒
k=1
Geometric
Series with −1 < x < 1
X
X
(−1)j(k) xk , −1 < x < 1, converge absolutely because
|x|k converges.
1
1
1
1
1
1
⇒ 1 − − 2 + 3 − 4 + 5 + 6 − · · · converge absolutely.
2 2
2
2
2
2
Conditional Convergence
Conditional
P Convergence
P
P
A series
ak is said to converge conditionally if
ak converges while
|ak |
diverges.
Alternating p-Series with 0 < p ≤ 1
X 1
X (−1)k
,
0
<
p
≤
1,
converge
conditionally
because
diverges.
kp
kp
∞
k+1
X
(−1)
1 1 1
= 1 − + − − · · · converge conditionally.
k
2 3 4
⇒
k=1
3
Alternating Series
Alternating Series
Alternating Series
Let {ak } be a sequence of positive numbers.
X
(−1)k ak = a0 − a1 + a2 − a3 + a4 − · · ·
is called an alternating series.
Alternating Series Test
Let {ak } be a decreasing sequence of positive numbers.
X
If ak → 0, then
(−1)k ak converges.
Alternating p-Series with p > 0
X (−1)k
1
p
, p > 0, converge since f (x) = p is decreasing, i.e., f 0 (x) = − p+1 >
kp
x
x
∞
X
(−1)k+1
1
1
1
0 for ∀x > 0, and lim f (x) = 0.
⇒
= 1 − + − − ···
x→∞
k
2
3
4
k=1
converge conditionally.
3
Examples
X (−1)k
1
2
, converge since f (x) =
is decreasing, i.e., f 0 (x) = −
>
2k + 1
2x + 1
(2x + 1)2
0 for ∀x > 0, and lim f (x) = 0.
x→∞
X (−1)k k
x
x2 − 10
, converge since f (x) = 2
is decreasing, i.e., f 0 (x) = − 2
>
2
k + 10√
x + 10
(x + 10)2
0, for ∀x > 10, and lim f (x) = 0.
x→∞
An Estimate for Alternating Series
An Estimate for Alternating Series
Let {ak } be a decreasing sequence of positive numbers that tends to 0 and let
∞
X
L =
(−1)k ak . Then the sum L lies between consecutive partial sums sn ,
sn+1 ,
k=0
sn < L < sn+1 , if n is odd; sn+1 < L < sn , if n is even.
and thus sn approximates L to within an+1
|L − sn | < an+1 .
Example
Find sn to approximate
∞
X
(−1)k+1
k=1
Set
∞
X
(−1)k+1
k=1
k
=
∞
X
(−1)k
k=0
k+1
k
=1−
1 1
+ · · · within 10−2 .
2 3
. For |L − sn | < 10−2 , we want
1
< 10−2 ⇒ n + 2 > 102 ⇒ n > 98.
(n + 1) + 1
Then n = 99 and the 99th partial sum s100 is
1 1 1
1
1
−
≈ 0.6882.
s99 = 1 − + − + · · · +
2 3 4
99 100
From the estimate
1
|L − s99 | < a100 =
≈ 0.00991.
101
we conclude that
∞
X
(−1)k+1
= ln 2 < 0.6981 ≈ s100
s99 ≈ 0.6882 <
k
an+1 =
k=1
4
Example
Find sn to approximate
∞
X
(−1)k+1
1
1
= 1 − + · · · within 10−2 .
(2k + 1)!
3! 5!
k=0
For |L − sn | < 10
−2
, we want
1
< 10−2 ⇒ n ≥ 1.
(2(n + 1) + 1)!
Then n = 1 and the 2nd partial sum s2 is
1
s1 = 1 −
≈ 0.8333
3!
1
From the estimate
|L − s1 | < a2 =
≈ 0.0083.
5!
we conclude that
∞
X
(−1)k+1
s1 ≈ 0.8333 <
= sin 1 < 0.8416 ≈ s2
(2k + 1)!
an+1 =
k=0
4
Rearrangements
Why Absolute Convergence Matters: Rearrangements (1)
Rearrangement of Absolute Convergence Series
∞
X
(−1)k
k=0
2k
=1−
1
1
1
1
1
2
+
− 3 + 4 − 5 + · · · = absolutely
2 22
2
2
2
3
Rearrangement 1 +
1
1
1
1
1
1
2
1
1
− + 4 + 6 − 3 + 8 + 10 − 5 · · · ? = =
2
2
2 2
2
2
2
2
2
3
Theorem 2. All rearrangements of an absolutely convergent series converge
absolutely to the same sum.
Why Absolute Convergence Matters: Rearrangements (2)
Rearrangement of Conditional Convergence Series
∞
X
(−1)k+1
k=1
k
=1−
Rearrangement 1 +
1 1 1 1 1
+ − + − + · · · = ln 2 conditionally
2 3 4 5 6
1
1
1 1 1 1 1 1
− + + − + +
− · · · ? = 6= ln 2
3 2 5 7 4 9 11 6
Multiply the original series by 12
∞
1 X (−1)k+1
1 1 1 1
1
1
= − + − +
+ · · · = ln 2
2
k
2 4 6 8 10
2
k=1
Adding the two series, we get the rearrangement
∞
∞
X
(−1)k+1
1 X (−1)k+1
1 1 1 1 1
3
+
= 1 + − + + − + · · · = ln 2
k
2
k
3 2 5 7 4
2
k=1
k=1
Remark
5
• A series that is only conditionally convergent can be rearranged to converge
to any number we please.
• It can also be arranged to diverge to +∞ or −∞, or even to oscillate
between any two bounds we choose.
Outline
Contents
1 Convergence Tests
1
2 Absolute Convergence
2.1 Absolute Convergence . . . . . . . . . . . . . . . . . . . . . . . .
2
2
3 Alternating Series
3
4 Rearrangements
5
6