Series Convergence - University of British Columbia

Series Convergence

Math 105 Elyse Yeager

Overview

A sequence is a list of numbers. A series is the "sum" of an infinite sequence, defined as the limit of the partial sums:

N

an := lim an. A convergent series is one in which that limit exists, and a divergent series is one in which that limit

N

n=a

n=1

does not exist. If a series has all positive terms, and is divergent, then the sum is going to infinity. However, in a series with

some terms negative, it is possible to be divergent without going to infinity.

Geometric

A geometric series is a series of the form {an} = {c ? rn} for constants c and r. Equivalently, a geometric series has terms that

differ by a constant ratio (hence the use of "r"): to get from one term to the next, you simply multiply by the constant r.

For any finite N ,

N

rn

=

1 - rN+1 .

In the infinite case, convergence depends on the absolute value of r. A large r will

1-r

n=0

cause divergence, a small r will cause convergence.

rn =

n=0

1 1-r

if |r| < 1

DIV if |r| 1

Divergence Test

If

{an}

is

a

series

and

lim

n

an

=

0,

then

an is divergent.

n=a

If

lim

n

an

=

0,

the

divergence

test

says

nothing,

and

we

need

another

test.

Integral Test

If a function f (x) is positive and decreasing, and we define a sequence {an} = {f (n)}, then same thing: they both converge, or both diverge.

n=a

an

and

a

f

(x)dx

do

the

p-test

Using the integral test, we can determine that, for a constant p,

n=1

1 np

is

convergent

when

p>1

and

divergent

when

p 1.

Comparison and Limit Comparison Tests

Suppose {an} and {bn} have positive terms. We know the convergence/divergence of one (say, by the integral test, or some other

test) and we suspect the two behave similarly, so we want to conclude something about the convergence/divergence of the other.

First, we may know that an bn for all n suitably large. In this case, if we know bn converges, then we can conclude an converges as well. If we know an diverges, then we can conclude bn diverges as well. However, if we know that an

converges, or if we know that bn diverges, we can't conclude anything. This is the Comparison Test, sometimes called the

Direct Comparison Test.

Second, we may suspect that the differences Comparison Test, which says that if lim an is a

n bn

between an and bn are not great when n is large. Then we use positive number (greater than zero, not infinite), then an and

the Limit bn do the

same thing: they both converge, or both diverge.

Ratio Test

Let

{an}

be

a

sequence

with

positive

terms.

We

consider

the

limit

lim

n

an+1 . an

If

the

limit

is

in

[0, 1),

then

the

series

converges.

If the limit is greater than 1 (including infinity), the series diverges. If the limit is equal to 1, the test is inconclusive, so we have

to try another test.

Absolute Convergence

If |an| converges, then it must be true that also an converges. This is helpful to know when you have a series with some negative terms, and you want to use a test that requires positive terms.

If |an| converges, we say an converges absolutely. If |an| diverges AND an converges, we say an converges conditionally.

Practice

Determine whether each of the series converges or diverges. If the series is geometric and convergent, give its value.

1.

1

k=1 k k + 1

2.

3(1.001)k

k=30

3. -1 n 5

n=3

4.

sin(n)

n=7

5.

cos(n)

n=7

6. ek k!

k=1

7. 2k 3k+2

k=0

8. 1 en

n=5

9. n!n! (2n)!

n=1

10. n2 + 1 2n4 + n

n=1

Solutions

Problem 1 We notice that this series is not geometric, and its terms tend to 0, so we can't evaluate it as a geometric series, and we can't use the divergence test.

To generate a guess about its convergence, we do the following:

1

1

1

1

=

=

k k+1

k2 + k

k2

k

We guess that our series behaves like the harmonic series, and the harmonic series diverges (which can be demonstrated by p-test or integral test). So, we guess that our series diverges. However, in order to directly compare our series to the harmonic series and show our series diverges, our terms would have to be BIGGER than the terms in the harmonic series, and this is not the case. So, we use limit comparison.

1 k 1 k2 +k

=

k2 + k k

=

k2 + k k2

=

1

lim

k

k 1

= lim

k

k2 +k

1 1+ =1

k

k2 + k k2 =

1 1 + , so

k

Since 1 is a real number greater than 0, by the Limit Comparison Test,

1 DIVERGES, like

k k+1

1 k

.

Problem 2 This is a geometric series with r = 1.001. Since |r| > 1, it is divergent.

Problem

3

This

is

a geometric

series with r =

-1 5

.

Since

|r| < 1,

it is convergent.

We want to use the formula

n=0

rn

=

1 1-r

,

but

our

series

does

not

start

at

0,

so

we

re-write

it:

-1 n -1 n 2 -1 n

1

11

=

-

=

- 1- +

5

5

5

1 - (-1/5)

5 25

n=3

n=0

n=0

1

1 1 5 -25 + 5 - 1

1

= -1+ - = +

=-

6/5

5 25 6

25

150

Problem 4 For any integer n, sin(n) = 0, so sin(n) = 0 = 0 . (So in particular, this series converges.)

Problem 5 For any integer n, cos(n) = ?1, so limn cos(n) = 0. By the divergence test, this series DIVERGES.

Problem 6 Factorials grow super fast. Like, wow, really fast. Even faster than exponentials. So the terms are going to zero, and the divergence test won't help us. Let's use ratio?it's a good go-to test with factorials.

ak+1 ak

=

ek+1 (k+1)!

ek k!

=

ek+1 ek

?

k! (k + 1)!

=e?

k(k - 1) ? ? ? (1) (k + 1)(k)(k - 1) ? ? ? (1)

=e?

1 k+1

=

e k+1

Since e is a constant,

lim ak+1 = lim e = 0

k ak

k k + 1

Since 0 < 1, by the ratio test, the series converges.

Problem 7 This is close to being in the form of a geometric series. First, we should have our powers be k, not k + 2, but we notice 3k+2 = 3k32 = 9 ? 32, so:

2k

2k

1 2k 1 2 k

3k+2 =

9 ? 3k = 9

3k = 9

3

k=0

k=0

k=0

k=0

now

it

looks

like

a

geometric

series

with

r

=

2 3

1

1

1

=

=

9 1 - (2/3)

3

In

conclusion:

this

(geometric)

series

is

convergent,

and

its

sum

is

1 3

.

Problem

8

Since

1 en

is

positive,

decreasing,

and

easily

integrable,

we

use

the

integral

test,

with

f (x) =

1 ex

= e-x.

e-x = lim -e-b - -e-5

5

b

= lim

b

11 e5 + eb

1 = e5

So, the integral converges. Then, by the integral test, the series converges as well.

Problem 9 Usually with factorials, we want to use the divergence test or the ratio test. Since the terms are indeed tending towards zero, we are left with the ratio test.

(n+1)!(n+1)!

an+1 = an

(2n+2)!

n!n! (2n)!

(n + 1)!(n + 1)! (2n)!

=

?

n!n!

(2n + 2)!

=

(n

+

1)&(n&)$(n$-$$1) ?

?

?

(&1) &

?

(n

+

1)&(n&)$(n$-$$1) ?

?

? (&1) &

?

?(2n?)$(2n$-$$1)$(2n$-$$2)

?

?

? (&1) &

&n$(n$-$$1) ? ? ?&(&1)

&n$(n$-$$1) ? ? ?&(&1)

(2n + 1)(2n + 1)?(2n?)$(2n$-$$1)$(2n$-$$2) ? ? ?&(&1)

1

= (n + 1)(n + 1) ?

, so

(2n + 2)(2n + 1)

lim an+1 = lim

(n + 1)(n + 1) = lim

(n + 1)(n + 1)

n+1 1

= lim

=

n an n (2n + 2)(2n + 1) n 2(n + 1)(2n + 1) n 4n + 2 4

Since the limit is a number less than 1, the series converges by the ratio test.

Problem 10 We want to make an estimation, when n gets big:

n2 + 1 n2

1

2n4 + n 2n4 = 2n2

Since

1 2n2

is

a convergent

series

(by p-test,

or

integral test),

we

guess that

our

series

is

convergent as

well.

If

we wanted

to

use

comparison

test,

we

should

have

to

show

n2 +1 2n4 +n

<

1 2n2

,

which

seems

unpleasant,

so

let's

use

limit

comparison.

lim

n

n2 +1 2n4 +n

1 2n2

(n2 + 1)2n2

= lim

n

2n4 + n

2n4 + 2n2

= lim

n

2n4 + n

1/n4 1/n4

=

lim

n

2 2

+ +

2 n2

1 n3

=1

Since the limit is a positive finite number, by the Limit Comparison Test, converges.

n1 +1 2n4 +n

does

the

same

thing

1 2n2

does:

it

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