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