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Math 55 chapter 20
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Convergence Tests(for Series with Positive and Negative Terms)
Math 55 - Elementary Anlaysis III
Institute of MathematicsUniversity of the Philippines
Diliman
Math 55 Convergence Tests II
Recall
R Divergence Test
If limn an 6= 0, then the series
n=1
an is divergent.
The following tests can be used for series with positive terms:
R Integral TestIf f is a positive, decreasing function continuous on [1,)such that an = f(n) n N, then the series
n=1
an
converges if
1
f(x) dx converges;
diverges if
1
f(x) dx diverges.
Math 55 Convergence Tests II
Recall
R Comparison Test
Given the seriesn=1
an andn=1
bn with positive terms,
if an bn andn=1
bn converges, thenn=1
an also converges.
if an bn andn=1
bn diverges, thenn=1
an also diverges.
R Limit Comparison Test
Given the seriesn=1
an andn=1
bn with positive terms, if
limn
anbn
= c > 0, then both series converge and both
diverge.
Math 55 Convergence Tests II
Alternating Series
Definition
An alternating series is a series whose terms are alternatelypositive and negative; it is of the form
n=1
(1)nbn orn=1
(1)n1bn
where bn > 0 for all n.
Examples.
1
n=1
(1)nn = 1 + 2 3 + 4 . . .
2
n=1
(1)n1n
= 1 12
+1
3 1
4+ called the alternating
harmonic series
Math 55 Convergence Tests II
Alternating Series
Theorem
If a given alternating series
n=1
(1)nbn orn=1
(1)n1bnsatisfies
(i) limn bn = 0;
(ii) bn+1 bn for all n, i.e., the sequence {bn} is decreasingthen it is convergent.
Math 55 Convergence Tests II
Alternating Series
Example
Show that the alternating harmonic series
n=1
(1)n1n
is
convergent.
Solution. We check the conditions of the Alternating SeriesTest. Let bn =
1n .
22 limn bn = limn
1
n= 0
22 bn+1 = 1n+ 1
1 and is therefore convergent.
Hence,n=1
(1)n1n2
is absolutely convergent.
Math 55 Convergence Tests II
More on Absolute Convergence
Theorem
Ifn=1
an is absolutely convergent, then it is convergent.
Proof. Supposen=1
an is absolutely convergent, i.e,n=1|an| is
convergent. For all n N,|an| an |an|
0 an + |an| 2|an|n=1
2|an| is convergent n=1
(an + |an|) is convergent. Hence,n=1
an =
n=1
(an + |an| |an|) =n=1
(an + |an|)n=1
|an|.
Since both series on the right-hand side of the equation are
convergent,n=1
an is convergent.
Math 55 Convergence Tests II
More on Absolute Convergence
Example
Show that the series
n=1
sinn
n2converges.
Solution. The given series has positive and negative terms but it isnot an alternating series so Alternating Series Test can not be used.
Consider the series
n=1
sinnn2. Note that sinnn2 = | sinn|n2 1n2
Since
n=1
1
n2converges,
n=1
sinnn2 also converges by Comparison
Test. Therefore,
n=1
sinn
n2is absolutely convergent, and hence,
convergent.Math 55 Convergence Tests II
More on Conditional Convergence
For conditionally convergent series, rearrangement of terms affects thesum. To illustrate, consider the alternating harmonic series and let
S =
n=1
(1)n1n
. The terms of the series can be rearranged as
S =(1 1
2 1
4
)+
(1
3 1
6 1
8
)+
(1
5 1
10 1
12
)+
=k=1
(1
2k 1 1
2(2k 1) 1
4k
)
=
k=1
(1
2(2k 1) 1
4k
)=
1
2
k=1
(1
(2k 1) 1
2k
)=
1
2
(1 1
2+
1
3 1
4+
)=
1
2S
Theorem (Riemann Series Theorem)
A conditionally convergent series may be made to converge toany desired value or to diverge by some rearrangement of terms.
Math 55 Convergence Tests II
Ratio Test
Theorem
Suppose
n=1
an is a series with nonzero terms.
(i) If limn
an+1an = L < 1, then the series is absolutely
convergent, and is therefore convergent.
(ii) If limn
an+1an = L > 1 or limn
an+1an =, then the series
is divergent.
Remark. The test is inconclusive if limn
an+1an = 1. In this
case, use another test!
Math 55 Convergence Tests II
Ratio Test
Example
Determine whether the series
n=1
nn
n!converges.
Solution. Let an =nn
n!. Then
limn
an+1an = limn (n+ 1)n+1(n+ 1)! n!nn = limn (n+ 1)(n+ 1)n(n+ 1)n! n!nn
= limn
(n+ 1
n
)n= lim
n
(1 +
1
n
)n= e > 1
Hence, the series diverges by Ratio Test.
Math 55 Convergence Tests II
Ratio Test
Example
Determine whether the series
n=1
(1)n1n2
2nconverges.
Solution. Let an = (1)n1n2
2n. Then
limn
an+1an = limn
(1)n (n+ 1)
2
2n+1
(1)n1n2
2n
= limn(n+ 1)2
2n+1 2
n
n2
= limn
1
2
(n+ 1
n
)2= lim
n1
2
(1 +
1
n
)2=
1
2< 1
Therefore, the series is absolutely convergent, and hence,convergent.
Math 55 Convergence Tests II
Root Test
Theorem
Given a seriesn=1
an,
(i) If limn |an|
1n = L < 1, then the series is absolutely
convergent, and is therefore convergent.
(ii) If limn |an|
1n = L > 1 or lim
n |an|1n =, then the series is
divergent.
Remark. The test is inconclusive if limn |an|
1n = 1. In this
case, use another test (but dont try Ratio Test because again,L = 1).
Math 55 Convergence Tests II
Root Test
Example
Determine whether the series
n=1
(2)nnn
converges or diverges.
Solution. Let an =(2)nnn
. Then
limn |an|
1n = lim
n
(2)nnn 1n
= limn
2
n= 0 < 1
Therefore, the series is absolutely convergent, and henceconvergent by Root Test.
Math 55 Convergence Tests II
Root Test
Example
Test the series
n=1
(2)n(tan1 n)n
for convergence.
Solution. Let an =(2)n
(tan1 n)n. Then
limn |an|
1n = lim
n
(2)n(tan1 n)n 1n
= limn
2
tan1 n=
2pi2
=4
pi> 1.
Hence the series diverges by Root Test.
Math 55 Convergence Tests II
Exercises
Determine whether the given series is absolutely convergent,conditionally convergent or divergent.
1
n=1
3 cosnn23 2
2
n=1
(1)n sin(
1
n
)3
n=2
(1)nlnn
4
n=1
cos(npi3
)n!
5
n=0
(1)n+1 nn3 + 5
6
n=1
(2)nn4
7
n=1
(1)nn2 2n
n!
8
n=0
2n+1
(n+ 2)2n
9
n=1
(1
5
)n(n2 + 1)
10
n=1
(1)n1 1 3 (2n 1)(2n 1)!
Math 55 Convergence Tests II
References
1 Stewart, J., Calculus, Early Transcendentals, 6 ed., ThomsonBrooks/Cole, 2008
2 Dawkins, P., Calculus 3, online notes available athttp://tutorial.math.lamar.edu/
Math 55 Convergence Tests II