11.2 Series In this section, we will learn about: Various types of series. INFINITE SEQUENCES AND SERIES

11.2

Series

In this section, we will learn about:

Various types of series.

INFINITE SEQUENCES AND SERIES

SERIES

If we try to add the terms of an infinite sequence

we get an expression of the form

a1 + a2 + a3 + ··· + an + ∙·∙

1{ }n na

Series 1

INFINITE SERIES

This is called an infinite series (or just a series).

It is denoted, for short, by the symbol

1

orn nn

a a

However, does it make sense to talk about the sum

of infinitely many terms?

INFINITE SERIES

It would be impossible to find a finite sum for the

series

1 + 2 + 3 + 4 + 5 + ∙∙∙ + n + ···

If we start adding the terms, we get the cumulative sums 1, 3, 6, 10, 15, 21, . . .

After the nth term, we get n(n + 1)/2, which becomes very large as n increases.

INFINITE SERIES

However, if we start to add the terms of the series

we get:

1 1 1 1 1 1 1

2 4 8 16 32 64 2n

3 7 15 31 6312 4 8 16 32 64 1 1/ 2, , , , , , , ,n

INFINITE SERIES

The table shows that, as we add more and more

terms, these partial sums become closer and closer

to 1.

In fact, by adding sufficiently many terms of the series, we

can make the partial sums as

close as we like to 1.

INFINITE SERIES

So, it seems reasonable to say that the sum of

this infinite series is 1 and to write:

1

1 1 1 1 1 11

2 2 4 8 16 2n nn

INFINITE SERIES

We use a similar idea to determine whether or not

a general series (Series 1) has a sum.

INFINITE SERIES

We consider the partial sums

s1 = a1

s2 = a1 + a2

s3 = a1 + a2 + a3

s3 = a1 + a2 + a3 + a4

In general,1 2 3

1

n

n n ii

s a a a a a

INFINITE SERIES

These partial sums form a new sequence {sn},

which may or may not have a limit.

If exists (as a finite number), then, as

in the preceding example, we call it the sum of

the infinite series an.

INFINITE SERIES

lim nn

s s

Given a series

let sn denote its nth partial sum:

1 2 31

ii

a a a a

1 21

n

n i ni

s a a a a

SUM OF INFINITE SERIES Definition 2

If the sequence {sn} is convergent and

exists as a real number, then the series ai is called

convergent and we write:

The number s is called the sum of the series.

Otherwise, the series is called divergent.

1 21

orn ii

a a a s a s

SUM OF INFINITE SERIES Definition 2

lim nn

s s

Thus, the sum of a series is the limit of the

sequence of partial sums.

So, when we write , we mean that, by adding

sufficiently many terms of the series, we can get as close as we like to the number s.

1n

n

a s

SUM OF INFINITE SERIES

Notice that:

1 1

limn

n in

n i

a a

SUM OF INFINITE SERIES

Compare with the improper integral

To find this integral, we integrate from 1 to t and then let t → .

For a series, we sum from 1 to n and then let n → .

1 1( ) lim ( )

t

tf x dx f x dx

SUM OF INFINITE SERIES VS. IMPROPER INTEGRALS

An important example of an infinite series is the

geometric series

2 3 1

1

1

0

n

n

n

a ar ar ar ar

ar a

GEOMETRIC SERIES Example 1

Each term is obtained from the preceding one by

multiplying it by the common ratio r.

We have already considered the special case where a = ½ and r = ½ earlier in the section.


If r = 1, then

sn = a + a + ∙∙∙ + a = na →

Since does not exist, the geometric series diverges

in this case.


lim nns

If r 1, we have

sn = a + ar + ar2 + ∙∙∙ + ar n–1

and

rsn = ar + ar2 + ∙∙∙ +ar n–1 + ar n


Subtracting these equations, we get:

sn – rsn = a – ar n

and(1 )

1

n

n

a rs

r

GEOMETRIC SERIES E. g. 1—Equation 3

If –1 < r < 1, we know from Result 9 in Section

11.1 that r n → 0 as n → .

So,

Thus, when |r | < 1, the series is convergent and its sum is a/(1 – r).

(1 )lim lim

1 1

n

nn n

a r as

r r


If r –1 or r > 1, the sequence {r n} is divergent by

Result 9 in Section 11.1

So, by Equation 3, does not exist.

Hence, the series diverges in those cases.


lim nn

s

The figure provides a

geometric demonstration

of the result in Example 1.

GEOMETRIC SERIES

If s is the sum of the series,

then, by similar triangles,

So,

s a

a a ar

GEOMETRIC SERIES

1

as

r

GEOMETRIC SERIES

We summarize the results of Example 1 as

follows.

The geometric series

is convergent if |r | < 1.

1 2

1

n

n

ar a ar ar

Result 4

Moreover, the sum of the series is:

If |r | 1, the series is divergent.

1

1

11

n

n

aar r

r

GEOMETRIC SERIES Result 4

Find the sum of the geometric series

The first term is a = 5 and the common ratio is r = –2/3

10 20 403 9 275


Since |r | = 2/3 < 1, the series is convergent by

Result 4 and its sum is:

23

53

10 20 40 55

3 9 27 1 ( )

5

3


What do we really mean when we say that the

sum of the series in Example 2 is 3?

Of course, we cannot literally add an infinite number of terms, one by one.

GEOMETRIC SERIES

However, according to Definition 2, the total

sum is the limit of the sequence of partial sums.

So, by taking the sum of sufficiently many terms, we can get as close as we like to the number 3.

GEOMETRIC SERIES

The table shows the first ten partial sums sn.The

graph shows how the sequence of partial sums

approaches 3.

GEOMETRIC SERIES

Is the series

convergent or divergent?

2 1

1

2 3n n

n


Let us rewrite the nth term of the series in the

form ar n-1:

We recognize this series as a geometric series with a = 4 and r = 4/3.

Since r > 1, the series diverges by Result 4.

12 1 2 ( 1)

11 1 1 1

4 42 3 (2 ) 3 4

3 3

nnn n n n

nn n n n


Write the number as a ratio

of integers.

After the first term, we have a geometric series with a = 17/103 and r = 1/102.

3 5 7

17 17 172.3171717 2.3

10 10 10


2.317 2.3171717...

Therefore,

3

2

17 1710 10002.317 2.3 2.3

1 991

10 10023 17

10 9901147

495


Find the sum of the series

where |x| < 1.

Notice that this series starts with n = 0.

So, the first term is x0 = 1.

With series, we adopt the convention that x0 = 1 even when x = 0.

0

n

n

x


Thus,

This is a geometric series with a = 1 and r = x.

2 3 4

0

1n

n

x x x x x


Since |r | = |x| < 1, it converges, and Result 4 gives:

0

1

1n

n

xx

GEOMETRIC SERIES E. g. 5—Equation 5

Show that the series

is convergent, and find its sum.

1

1

( 1)n n n

SERIES Example 6

This is not a geometric series.

So, we go back to the definition of a convergent series and compute the partial sums:

1

1

( 1)

1 1 1 1

1 2 2 3 3 4 ( 1)

n

ni

si i

n n

SERIES Example 6

We can simplify this expression if we use the

partial fraction decomposition.

See Section 7.4

1 1 1

( 1) 1i i i i

SERIES Example 6

Thus, we have:

1

1

1

( 1)

1 1

1

1 1 1 1 1 1 11

2 2 3 3 4 1

11

1

n

ni

n

i

si i

i i

n n

n

SERIES Example 6

Thus,

Hence, the given series is convergent and

1lim lim 1 1 0 1

1nn n

sn

1

11

( 1)n n n

SERIES Example 6

The figure illustrates Example 6 by showing the

graphs of the sequence of terms an =1/[n(n + 1)]

and the sequence {sn} of partial sums.

Notice that an → 0 and sn → 1.

SERIES

Show that the harmonic series

is divergent.

1

1 1 1 11

2 3 4n n

HARMONIC SERIES Example 7

For this particular series it’s convenient to consider

the partial sums s2, s4, s8, s16, s32, …and show that

they become large.

1

12 2

1 1 1 1 1 14 2 3 4 2 4 4

2

2

1

1

1 1

1

s

s

s


1 1 1 1 1 1 18 2 3 4 5 6 7 8

1 1 1 1 1 1 1

2 4 4 8 8 8 8

1 1 1

2 2 2

3

2

1

1

1

1

s


Similarly,

1 1 1 1 1 1 116 2 3 4 5 8 9 16

1 1 1 1 1 1 1

2 4 4 8 8 16 16

1 1 1 1

2 2 2 2

4

2

1

1

1

1

s


Similarly,

Similarly, s32 > 1 + 5/2, s64 > 1 + 6/2, and, in

general,

This shows that s2n → as n → , and so {sn} is divergent.

Therefore, the harmonic series diverges.


21

2n

ns

HARMONIC SERIES

The method used in Example 7 for showing that

the harmonic series diverges is due to the French

scholar Nicole Oresme (1323–1382).

If the series is convergent, then1

nn

a

lim 0n

na

SERIES Theorem 6

Let sn = a1 + a2 + ∙∙∙ + an

Then, an = sn – sn–1

Since an is convergent, the sequence {sn} is convergent.

SERIES Theorem 6 - Proof

Let

Since n – 1 → as n → , we also have:


lim nn

s s

1lim nn

s s

Therefore,

1

1

lim lim

lim lim

0

n n nn n

n nn n

a s s

s s

s s


With any series an we associate two sequences:

The sequence {sn} of its partial sums

The sequence {an} of its terms

SERIES Note 1

If an is convergent, then

The limit of the sequence {sn} is s (the sum of the series).

The limit of the sequence {an}, as Theorem 6 asserts, is 0.

SERIES Note 1

The converse of Theorem 6 is not true in general.

If , we cannot conclude that an is

convergent.

SERIES Note 2

lim 0nn

a

Observe that, for the harmonic series 1/n, we

have an = 1/n → 0 as n → ∞.

However, we showed in Example 7 that 1/n is divergent.

SERIES Note 2

If does not exist or if ,

then the series is divergent.

lim nna

lim 0nn

a

1n

n

a

THE TEST FOR DIVERGENCE Test 7

The Test for Divergence follows from Theorem 6.

If the series is not divergent, then it is convergent.

Thus,

TEST FOR DIVERGENCE

lim 0nn

a

Show that the series diverges.

So, the series diverges by the Test for Divergence.

2

21 5 4n

n

n

TEST FOR DIVERGENCE Example 8

2

2 2

1 1lim lim lim 0

5 4 5 4 / 5nn n n

na

n n

If we find that , we know that an

is divergent.

If we find that , we know nothing about

the convergence or divergence of an.

SERIES Note 3

lim 0nna

lim 0nn

a

Remember the warning in Note 2:

If , the series an might converge or

diverge.

SERIES

lim 0nn

a

Note 3

If an and bn are convergent series, then so are the

series can (where c is a constant), (an + bn), and

(an – bn), and

1 1

1 1 1

1 1 1

i.

ii.

iii.

n nn n

n n n nn n n

n n n nn n n

ca c a

a b a b

a b a b

SERIES Theorem 8

These properties of convergent series follow from

the corresponding Limit Laws for Sequences in

Section 11.1

For instance, we prove part ii of Theorem 8 as follows.

SERIES

Let

1 1

1 1

n

n i ni n

n

n i ni n

s a s a

t b t b

THEOREM 8 ii—PROOF

The nth partial sum for the series Σ (an + bn) is:

1

n

n iii

u a b


Using Equation 10 in Section 5.2, we have:

1

1 1

1 1

lim lim

lim

lim lim

lim lim

n

n i in n

i

n n

i in

i i

n n

i in n

i i

n nn n

u a b

a b

a b

s t s t


Hence, Σ (an + bn) is convergent, and its sum is:

1

1 1

n nn

n nn n

a b s t

a b


Find the sum of the series

The series 1/2n is a geometric series with a = ½ and

r = ½. Hence,

1

3 1

( 1) 2nn n n

SERIES Example 9

12

11 2

11

2 1nn

In Example 6, we found that:

So, by Theorem 8, the given series is convergent

and

1 1 1

3 1 1 13

( 1) 2 ( 1) 2

3 1 1

4

n nn n nn n n n

SERIES Example 9

1

11

( 1)n n n

A finite number of terms does not affect the

convergence or divergence of a series.

SERIES Note 4

For instance, suppose that we were able to show

that the series is convergent.

Since

it follows that the entire series is convergent.

34 1n

n

n

SERIES Note 4

3 31 4

1 2 3

1 2 9 28 1n n

n n

n n

31 1n

n

n

Similarly, if it is known that the series

converges, then the full series

is also convergent.

1n

n N

a

1 1 1

N

n n nn n n N

a a a

SERIES Note 4

Documents

11.2 Series In this section, we will learn about: Various types of series. INFINITE SEQUENCES AND SERIES