infinite series tests - saint joseph's...

Infinite Series Tests Convergent or Divergent? First, it is important to recall some key definitions and notations before discussing the individual series tests. sequence - a function that takes on real number values a

n and is defined on the set of positive

integers n = 1, 2, 3, 4, ... Listed Notation -

an{ } = a1,a2 ,a3,K,an,K

where for example 1n

na

n=

+ n = 1, 2, 3, 4, ... series - a summation of a sequence of numbers Notation -

an! = a

1+ a

2+ a

3+K+a

n+K

convergence - whenever a sequence or a series has a limit divergence - whenever a sequence or a series does not have a limit GEOMETRIC SERIES The first type of series test is used only for a geometric series. A geometric series is a series which follows the pattern, a + ar + ar

2

+K+arn

+K

where a is the initial term and r is a ratio (i.e., 1

2, 3

4, 5

3, etc. ).

If r < 1 (that is, !1 < r < 1), then the series converges.

further, this series converges to a

1 ! r

Example:

Determine whether the series

2 +1 +1

2+1

4+K converges.

Solution: To find r divide any term by the term preceding it.

The series follows the pattern of a geometric series with a = 2 and r =1

2 .

By the test for convergence, this series does converge, since 1

2=1

2<1 .

It follows that the series converges to 4 since

2 2 24

11 1 0.5 0.512

a

r= = = =

! !!

Adapted from http://www.utexas.edu/student/utlc/lrnres/mstc/a315/calchandouts/infiniteseries.doc 2

POSITIVE SERIES An infinite series with no negative terms is often referred to as a positive series. Positive series are usually denoted,

a1+ a

2+ a

3+K+a

n+K

There are several tests used to determine whether or not a positive series converges or diverges. THE DIVERGENCE TEST (Section 11.2): If lim

n!"an# 0 , then the series diverges. Otherwise, the test is inconclusive.

This test is useful in quickly determining whether a given series diverges. It asks the question, "Is the series getting continually larger?" If it continues to grow, then the series diverges.

Example:

Does the following series converge or diverge: 2n

n=1

!

"

Solution: Examine the first couple of terms. Does the series continue to grow? 2 + 4 + 6+K The series is getting continually larger. Thus lim

n!"an= " # 0 and the series diverges.

THE INTEGRAL TEST (Section 11.3): Let f (x) > 0 for x ! 1 , and f (x) is a continuous decreasing function. Given an = f (n) ,

if f (x)dx1

!

" is convergent, then so is 1

n

n

a

!

=

" ;

if f (x)dx1

!

" is divergent, then so is 1

n

n

a

!

=

" .

This test is useful for the inverse trig functions, as well as determining the convergence or divergence of a particularly important series, the harmonic series, which is the series in the example below. Example: Determine whether or not the series converges.

1 +1

2+1

3+K+

1

n+K

Solution: The nth-term test doesn’t help here since lim

n!"an= 0 (the terms are getting smaller).

Now f (x) =1

x is a continuous, positive, decreasing function. Using the integral test

1 1

1

1 1lim

lim(ln ln1)

lim ln

0

b

b

b

b

b

dx dxx x

b

x

!

"!

"!

"!

=

= #

=

= ! #

= !

$ $

Since the integral is divergent, then the series is divergent.


P-SERIES TEST (Section 11.3):

If 0 < p ! 1 , then the series 1

np

n=1

!

" diverges.

If p > 1, then the series

1

np

n=1

!

" converges.

This test is derived from the integral test and is useful only for the series which contains the np term. However, it will be quite helpful in following comparison tests. COMPARISON TEST (Section 11.4): Given three series, pn , d

n, c

n:

if 0 ! pn ! cn for each n and cn

n=1

!

" converges, then pnn=1

!

" converges;

if 0 ! dn ! pn for each n and dn

n=1

!

" diverges, then pnn=1

!

" diverges.

This test is used quite often and is one of the more important ones to know. It requires familiarity with the convergence or divergence of other series, in addition to being able to compare them with a given series. This test proves very useful when a series appears “un-testable.” Example #1:

Does the series n + 2

n + 3n=1

!

" #1

n3

converge or diverge?

Solution:

This series closely resembles the p-series, 1

n3

, which converges. In order to

compare the two series using the comparison test, determine which series is larger.

Since 21

3

n

n

+<

+ for any n,

n + 2

n + 3!1

n3<1

n3

. (Multiplying each side by 1

n3

preserves the inequality since 1

n3

is a positive term for all n greater than 1) So

by the comparison test, n + 2

n + 3n=1

!

" #1

n3

converges.

Example #2:

Does the series n + 5

n2+ 3nn=1

!

" converge or diverge?

Solution:

Factor the denominator, so the series can be represented as n + 5

n(n + 3)n=1

!

"


Then, split up the fraction into n + 5

n + 3n=1

!

" #1

n . Note that

51

3

n

n

+>

+ , so

5 1 1

3

n

n n n

+! >

+. Thus, by comparing this series with the divergent harmonic

series, n + 5

n2+ 3nn=1

!

" diverges.

LIMIT COMPARISON TEST (Section 11.4): Given three series, pn , d

n, c

n:

if limn!"

pn

cn exists, and c

n

n=1

!

" is a convergent series, then pnn=1

!

" converges;

if limn!"

pn

dn exists and is not equal to 0 or ! , and d

n

n=1

!

" is a divergent series,

then pnn=1

!

" diverges.

Similar to the Comparison Test, this test is also helpful when confronted with a series that appears un-testable. Most often, this test can be applied to fractions with polynomials or with a combination of polynomials and exponentials. Example:

Determine the convergence or divergence of the series n2 + 3n + 7

4n4+ 3n

2+10n=1

!

"

Solution: As n increases, the highest powers of the numerator and the denominator become

the significant terms, in this case n2 and 4n4 . So n2

4n4

is similar to 1

4n2

,

which is a convergent p-series. Dividing the series by 1

4n2

(i.e., multiply by

the reciprocal ) results in 4n

4 +12n3 + 28n2

4n4+ 3n

2+10

.

Now 4 3 2

4 2

4 12 28 4lim 1

4 3 10 4n

n n n

n n!"

+ += =

+ +. So, since the limit exists,

the series is a convergent series. SERIES WITH BOTH POSITIVE AND NEGATIVE TERMS

The next sets of series tests are those that apply to series with both negative and positive terms (known as alternating series). All the previous tests apply only to those tests whose terms are all positive. An alternating series is a series of the form

(!1)n+1an= a1 ! a2 + a3 ! a4 + a5 !L

n=1

"

# ,

in other words, a series whose terms alternate between positive and negative. Note that a

n is a sequence of positive

numbers and the (!1)n +1 determines the sign.


ALTERNATING SERIES TEST (Section 11.5): For a given alternating series, consider a

n (the sequence of positive numbers without the (!1)n +1)

if an+1

< an

and limn!"

an= 0 , then the series converges.

This test will immediately give us the convergence of an alternating series, just by looking at its basic behavior. If each number gets progressively smaller, and the series approaches 0, then the series converges.

Example:

Determine whether the series 1

( 1)n

n n

!

=

"# converges or diverges.

Solution:

For this series, the terms continue to decrease and limn!"

1

n= 0 and

because these 2 criterion are met, the series converges. ABSOLUTE-CONVERGENCE TEST (Section 11.6):

If an

n=1

!

" converges, then so does an

n=1

!

" .

Like the alternating series test, this test checks only for the convergence of a series and says nothing about the divergence.

Example:

Does the series (!1)n2n

(n + 2)2

n=1

"

# converge?

Solution:

Take the absolute value of the series, and get 2n

(n + 2)2

n=1

!

" . By

applying the divergence test, this series diverges (the series continues to grow), so the absolute-convergence test is inconclusive. A later test might help solve this problem. A few definitions will help clarify certain types of convergence. A convergent series a

n! is said to converge absolutely if an! converges as well.

A convergent series a

n! is said to converge conditionally if an! diverges.

For example, the alternating harmonic series converges conditionally. RATIO TEST (Section 11.6):

Given a series an

n=1

!

" , take the limit limn!"

an+1

an

= L

if L <1 , then the series converges; if L >1 or ! , then the series diverges.

This test says nothing of the series if this limit equals one. The ratio test can be used in factorials and exponentials

UT Learning Center Jester A332A • 512-471-3614 (846) 12/99 University of Texas at Austin

Example:

Determine the convergence or divergence of the series !1

4

" #

$ %

n

n=1

&

' .

Solution:

Applying the ratio test, we get limn!"

1

4n+1

1

4n

=4n

4n+1 =

1

4< 1. Thus the series converges. (Of

course, the geometric series test would also work here.

Example:

Determine the convergence for the series, 3n

n!n=1

!

" .

Solution: This series has both a factorial and an exponent. First, it is important to

determine the n + 1st term, which is

!

an+1 =

3n+1

(n +1)!

Now, place it in the ratio test format, such that

!

an+1

an

=

3n+1

(n +1)!

3n

n!

=3n!

(n +1)!=

3

n +1 .

So, limn!"

3

n +1= 0 < 1, thus the series converges.

ROOT TEST (Section 11.6): For a given series:

if

!

limn"#

an

n = L <1, then the series converges (in fact, it’s absolutely convergent);

if

!

limn"#

an

n >1 or

!

limn"#

an

n =#, then the series diverges;

if

!

limn"#

an

n =1, then the test is inconclusive.

This test will assist in determining the convergence or divergence of monomials of degree n, especially when the term nn appears in the series. The only terms that appear in such a series should be various degrees of n alone and no other terms—this test’s effectiveness is limited to a small array of series.

Example:

Determine the convergence of the series 2n2

nn

n=1

!

"

Solution: Since the series has both exponentials in the top and bottom, the

root test should be applied. So,

2

2 2n n

n

nn n

= and limn!"

2n

n=" .

Thus, by the root test, this series diverges.

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