Section 11.5 Lecture Slides

1. Chapter 11: Sequences, Series, and Power Series
Section 11.5: Alternating Series and Absolute Convergence
Alea Wittig
SUNY Albany

2. Outline
Alternating Series
Absolute Convergence and Conditional Convergence

3. ▶ So far we have looked at methods to deal with series whose
terms are positive.
▶ Here we deal with series where terms are not necessarily
positive.

4. Alternating Series

5. Alternating Series
An alternating series is one whose terms are alternately positive
and negative.
1 −
1
2
+
1
3
−
1
4
+
1
5
−
1
6
+ . . . =
X
n=1
(−1)n−1 1
n
−
1
2
+
2
3
−
3
4
+
4
5
−
5
6
+
6
7
− . . . =
X
n=1
(−1)n n
n + 1
In these examples we have
an = (−1)n
bn or an = (−1)n−1
bn
where bn is a positive number. (bn = |an|.)

6. Alternating Series Test
Theorem: If the alternating series
∞
X
n=1
(−1)n−1
bn = b1 − b2 + b3 − b4 + . . . (bn > 0)
satisfies the conditions
(i) bn+1 ≤ bn for all n, ie, bn is decreasing.
(ii) lim
n→∞
bn = 0
then the series is convergent

7. Alternating Series Test

8. Remarks
▶
P
an =
P
(−1)n−1bn or
P
(−1)nbn where bn > 0, means
bn = |an|
▶ If (ii) is false then
P
an diverges:
lim
n→∞
|an| ̸= 0 =⇒ lim
n→∞
an ̸= 0
by Theorem 6 of the 11.1 of the lecture notes
The Test for Divergence: lim
n→∞
an ̸= 0 =⇒
∞
X
n=1
an diverges.

9. Example 1
1/2
Determine whether the alternating harmonic series
X
n=1
(−1)n−1
n
converges.

10. Example 1
2/2
(i) bn+1 ≤ bn:
bn =
1
n
bn+1 ≤ bn ⇐⇒
1
n + 1
≤
1
n
✓
(ii) limn→∞ bn = 0:
lim
n→∞
bn = lim
n→∞
1
n
= 0✓
Thus the alternating harmonic series converges by AST.

11. Example 2
1/2
Determine whether the series
∞
X
n=1
(−1)n 3n
4n − 1
converges.

12. Example 2
2/2
∞
X
n=1
(−1)n 3n
4n − 1
is alternating with bn = 3n
4n−1
(i) bn+1 ≤ bn
3n + 3
4n − 3
≤
3n
4n − 1
⇐⇒
(3n + 3)(4n − 1) ≤ 3n(4n − 3) ⇐⇒
12n2
− 3n + 12n − 3 ≤ 12n2
− 9n ⇐⇒
12n2
− 9n − 3 ≤ 12n2
− 9n✓
(ii) lim
n→∞
bn = 0:
lim
n→∞
3n
4n − 1
=
3
4
̸= 0 ×
Fails (ii) so, as explained in remarks, by the test for divergence the
series diverges.

13. Example 3
1/3
Test the series
∞
X
n=1
(−1)n+1 n2
n3 + 1
for convergence or divergence.

14. Example 3
2/3
bn =
n2
n3 + 1
(i) We can show that bn is decreasing by showing that
f (x) = x2
x3+1
is decreasing by taking the derivative:
f ′
(x) =
(x3 + 1)(2x) − x2(3x2)
(x3 + 1)2
< 0 ⇐⇒
(x3
+ 1)(2x) − x2
(3x2
) < 0 ⇐⇒ 2x4
+ 2x − 3x4
< 0 ⇐⇒
2x − x4
< 0 ⇐⇒ 2 < x3
So bn is decreasing for n ≥ 2 ✓.
(ii) lim
n→∞
n2
n3 + 1
= 0✓
So the series is convergent by AST.

15. Absolute Convergence and Conditional
Convergence

16. ▶ Given any series
P
an, we can consider the corresponding
series
∞
X
n=1
|an| = |a1| + |a2| + |a3| + . . .
eg
P
an =
P
(−3)n
=⇒
P
|an| =
P
3n
eg
P
an =
P (−1)n
n =⇒
P
an =
P 1
n

17. Absolute Convergence
Definition: A series
P
an is called absolutely convergent if
the series of absolute values
P
|an| is convergent.
Theorem: If a series
P
an is absolutely convergent, then it is
convergent.

18. Example 4
1/2
Determine whether the series
∞
X
n=1
(−1)n−1
n2
= 1 −
1
22
+
1
32
−
1
42
+ . . .
converges.

19. Example 4
2/2
X
|an| =
X 1
n2
is a p series with p = 2 > 1 which converges
=⇒
X
an =
X (−1)n−1
n2
is absolutely convergent thus convergent

20. Example 5
1/2
Determine whether the series
∞
X
n=1
cos n
n2
=
cos (1)
12
+
cos (2)
22
+ . . .
is convergent or divergent

21. Example 5
2/2
▶
P cos n
n2 is an example a series which is positive for some values
of n and negative for others, but is not of the form
P
(−1)nbn.
▶ We use the fact that
−1 ≤ cos θ ≤ 1 for all θ
to determine if the series is absolutely convergent.
0 ≤ |an| =
| cos n|
n2
≤
1
n2
▶ Since
P 1
n2 is convergent,
P
|an| is convergent by direct
comparison.
▶ Thus
P
an =
P cos n
n2 is absolutely convergent, and thus
convergent.

22. Conditional Convergence
Definition A series
P
an is called conditionally convergent
if it is convergent but not absolutely convergent; that is,
P
an
converges but
P
|an| diverges.
eg, The alternating harmonic series is conditionally convergent.
∞
X
n=1
(−1)n
n
converges by AST whereas
∞
X
n=1
1
n
diverges

23. Example 6
1/4
Determine whether the series is absolutely convergent, conditionally
convergent, or divergent.
(a)
∞
X
n=1
(−1)n
n3
(b)
∞
X
n=1
(−1)n
3
√
n
(c)
∞
X
n=1
(−1)n n
2n + 1

24. Example 6
2/4
(a)
∞
X
n=1
(−1)n
n3
|an| =
1
n3
=⇒
∞
X
n=1
|an| =
∞
X
n=1
1
n3
← convergent p series p = 3
=⇒
∞
X
n=1
(−1)n
n3
is absolutely convergent.

25. Example 6
3/4
(b)
∞
X
n=1
(−1)n
3
√
n
X
|an| =
X 1
3
√
n
is a divergent p-series with p =
1
3
≤ 1
▶ This means that
X
an is not absolutely convergent.
▶ Check for conditional convergence using AST:
(i) lim
n→∞
bn = lim
n→∞
1
3
√
n
= 0 ✓
(ii) bn+1 ≤ bn ⇐⇒
1
3
√
n + 1
≤
1
3
√
n
⇐⇒ 1 ≤
3
r
n + 1
n
⇐⇒ 1 ≤
n + 1
n
= 1 +
1
n
✓
So
∞
X
n=1
(−1)n
3
√
n
is conditionally convergent.

26. Example 6
4/4
(c)
∞
X
n=1
(−1)n n
2n + 1
lim
n→∞
bn = lim
n→∞
n
2n + 1
=
1
2
̸= 0
=⇒
∞
X
n=1
(−1)n n
2n + 1
is divergent by test for divergence.