INFINITE SEQUENCES AND SERIES
8.4 Other Convergence Tests
Objective: Determine if a series is convergent or divergent
I. This section deals with series whose terms are not necessarily positive.
II. Alternating series test:
A. If
∞
∑ ( −1)
n− 1
bn = b1 − b2 + b 3 − b 4 + . . . with bn > 0 satisfies
n =1
1. bn+ 1 ≤ bn for all n, and
2.
lim bn = 0, then the series is convergent.
n→ ∞
B. If the terms of an alternating series decrease to 0 in absolute value, then the series
is convergent.
III. The alternating harmonic series:
A. n + 1 > n ⇒
∞
(− 1)n −1
1
, bn =
> 0
n
n
n =1
∑
1
1
< ⇒ bn+ 1 ≤ bn for all n
n+1 n
1
=0
n→ ∞ n
C. The series converges by the Alternating Series Test.
B.
lim
∞
( − 1)n 3n
IV. Test
for convergence
4n − 1
n =1
∑
A.
lim bn = lim
n→ ∞
n→ ∞
3n
3
= , so the alternating series test does not apply.
4n − 1 4
( − 1)n 3n
does not exist, the series diverges by the test for
n → ∞ 4n − 1
B. Since lim an = lim
n→ ∞
divergence.
V. The error involved in using the partial sum s n as an approximation to the total sum s is the
remainder Rn = s − sn .
VI. Alternating Series Estimation Theorem: If s =
∞
∑ ( −1)
n =1
series that satisfies
n− 1
bn is the sum of an alternating
A. 0 < bn+ 1 ≤ bn and lim bn = 0, then | R n | = | s − s n | ≤ bn+ 1 .
n→ ∞
B. The error (in using sn to approximate s) is smaller than the first neglected term.
[only applies to alternating series]
VII. Find the sum of
B.
C.
E.
( −1)n
correct to three decimal places.
n!
i=0
∑
1
[0! = 1, by definition]
> 0 for all n
n!
1
1
1
=
<
(n + 1)!
n!(n + 1) n!
1
1
1
1
0<
< ; lim
= 0 ⇒ lim
=0
n → ∞ n!
n! n n → ∞ n
The series is convergent by the alternating series test.
1
1
1
1
−
+
−
+...
s =
0! 1! 2! 3!
A.
D.
∞
1
1
<
= 0.0002
5040 5000
2. s 6 ≈ 0.368056 ⇒ | s − s 6 | ≤ b 7 < 0.0002 ⇒ s ≈ 0.368 correct to three
decimal places.
1. b7 =
VIII. A series
∑a
n
is absolutely convergent if the series of absolute values
convergent.
A.
∞
( − 1)n − 1
is absolutely convergent because
2
n
n =1
∑
∞
∑
n =1
(− 1)n − 1
=
n2
∞
1
∑n
n =1
2
∑| a
n
| is
is convergent
[p-series with p = 2].
B. The alternating harmonic series
∞
convergent because
∑
n =1
∞
( − 1)n − 1
is convergent, but not absolutely
n
n =1
∑
(− 1)n − 1
=
n
∞
1
∑n
which is the divergent harmonic series.
n =1
[p-series with p = 1]
IX. If a series
X. Test
∞
∑
n =1
∑a
n
is absolutely convergent, then it is convergent.
cosn
for convergence
n2
[not reversible!]
A.
∞
∑
n =1
cosn
cos1 cos2 cos3
=
+
+
. . . has both positive and negative terms, but
2
n
12
22
32
they do not alternate.
∞
1.
∑
n =1
2.
cosn
=
n2
cosn
2
≤
∞
∑
n =1
cosn
n2
1
n2
1
3. Since
is convergent [p-series with p = 2],
n2
the comparison test.
n
∑
B. Therefore, the given series
∑
cosn
n2
∑
cosn
n2
is convergent by
is absolutely convergent, which means it is
also convergent.
XI. The ratio test
A. If lim
n→ ∞
an +1
an
∞
= L < 1, then the series
∑a
n
is absolutely convergent (and
n =1
therefore convergent).
a
a
B. If lim n +1 = L > 1 or lim n +1 = ∞ , then the series
n → ∞ an
n → ∞ an
C. If lim
n→ ∞
X. Test
∞
∑
( − 1)n
n =1
an +1
an
∞
∑a
n
is divergent.
n =1
= 1, the ratio test fails.
n3
for absolute convergence
3n
n3
3n
(− 1)n +1(n + 1)3
3
3
(n + 1)3 3n
1  n + 1
1
1
1
3n +1
⋅
=
1
+
→
<1
=
=
=




n 3
n +1
3
3 n 
3
n
3
( − 1) n
3
n
3n
A. an = (− 1)n
B.
an+ 1
an
C. By the ratio test, the given series is absolutely convergent and therefore
convergent.