From last class: The ratio test vs the integral test
P
The ratio test is useful for series k ak where ak involves terms of
the form ak or k!, e.g.,
∞
X
(k!)3
(3k)!
k=1
for which the integral test does not provide a solution (why?). It is
however less useful if ak involves polynomials in k, such as
∞
X
1
.
kp
k=1
Here the ratio test is inconclusive, but the integral test works!
Math 105 (Section 204)
Series: Tests of convergence
2011W T2
1/5
Comparison tests
Basic comparison test
Let
P
ak and
P
bk be series with positive terms.
P
P
1. If 0 < ak ≤ bk and
bk converges, then
ak converges.
P
P
2. If 0 < bk ≤ ak and
bk diverges, then
ak diverges.
k
k
Limit Comparison Test
Let
P
k
ak and
P
k
bk be series with positive terms, and
ak
= L.
k→∞ bk
P
P
1. If 0 < L < ∞, then
ak and
bk both converge or both diverge.
P
P
2. If L = 0 and
bk converges, then
ak converges.
P
P
3. If L = ∞ and
bk diverges, then
ak diverges.
lim
Math 105 (Section 204)
Series: Tests of convergence
2011W T2
2/5
Test your testing skills
Which of the following series converge?
∞
X
k=1
1
k− k ,
∞
X
2k k!
k=1
kk
,
∞
X
k=1
tan
1
k
1. first
2. second
3. third
4. first and second
5. first and third
Math 105 (Section 204)
Series: Tests of convergence
2011W T2
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Absolute and conditional convergence
Definition
P
Let k ak be an infinite series whose summands are not necessarily of the
same sign.
P
P
The series k ak is said to be absolutely convergent if k |ak |
converges.
P
P
The series k P
ak is said to be conditionally convergent if k ak
converges but k |ak | does not.
Math 105 (Section 204)
Series: Tests of convergence
2011W T2
4/5
Examples of absolutely and conditionally convergent series
The series
∞
X
(−1)n
n=1
n2
= −1 +
1 1 1
− + − ···
2 3 4
converges absolutely by the p-series test.
The series
1−
1 1 1 1 1
+ − + − + ···
2 2 3 3 4
converges conditionally.
Math 105 (Section 204)
Series: Tests of convergence
2011W T2
5/5