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

11INFINITE SEQUENCES AND SERIESINFINITE 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.

INFINITE SERIES

If exists (as a finite number),

then, as in the preceding example,

we call it the sum of the infinite series Σ an.

lim nns s

SUM OF INFINITE SERIES

Given a series

let sn denote its nth partial sum:

1 2 31n

na 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 Σ an 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 nn

a a a s a s

SUM OF INFINITE SERIES Definition 2

lim nns 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.

GEOMETRIC SERIES Example 1

If r = 1, then

sn = a + a + ∙∙∙ + a = na → ±∞

Since doesn’t exist, the geometric series diverges in this case.

GEOMETRIC SERIES Example 1

lim nns

If r ≠ 1, we have

sn = a + ar + ar2 + ∙∙∙ + ar n–1

and

rsn = ar + ar2 + ∙∙∙ +ar n–1 + ar n

GEOMETRIC SERIES Example 1

Subtracting these equations,

we get:

sn – rsn = a – ar n

(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 lim

1 1 1 1

nn

nn n n

a r a a as r

r r r r

GEOMETRIC SERIES Example 1

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.

GEOMETRIC SERIES Example 1

lim nns

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

GEOMETRIC SERIES Result 4

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

GEOMETRIC SERIES Example 2

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

GEOMETRIC SERIES Example 2

What do we really mean when we say

that the sum of the series in Example 2

is 3?

Of course, we can’t 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

GEOMETRIC SERIES Example 3

Let’s 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

GEOMETRIC SERIES Example 3

Write the number

as a ratio of integers.

2.3171717…

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

3 5 7

17 17 172.3

10 10 10

GEOMETRIC SERIES Example 4

2.317 2.3171717...

Therefore,

3

2

17 1710 10002.317 2.3 2.31 99

110 100

23 17

10 9901147

495

GEOMETRIC SERIES Example 4

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

GEOMETRIC SERIES Example 5

Thus,

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

2 3 4

0

1n

n

x x x x x

GEOMETRIC SERIES Example 5

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 1 112 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 convergentand

1lim lim 1 1 0 1

1nn ns

n

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 1 112 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

HARMONIC SERIES Example 7

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

HARMONIC SERIES Example 7

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

HARMONIC SERIES Example 7

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.

HARMONIC SERIES Example 7

212

n

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,

then

1n

n

a

lim 0nna

SERIES Theorem 6

Let sn = a1 + a2 + ∙∙∙ + an

Then, an = sn – sn–1

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

SERIES Theorem 6—Proof

Let

Since n – 1 → ∞ as n → ∞,

we also have:

SERIES Theorem 6—Proof

lim nns s

1lim nns s

Therefore,

11

lim lim

lim lim

0

n n nn n

n nn n

a s s

s s

s s

SERIES Theorem 6—Proof

With any series Σ an we associate

two sequences:

The sequence {sn} of its partial sums

The sequence {an} of its terms

SERIES Note 1

If Σ 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 Σ an is convergent.

SERIES Note 2

lim 0nna

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

we have an = 1/n → 0 as n → ∞.

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

SERIES Note 2

If does not exist or if ,

then the series

is divergent.

lim nna

lim 0nna

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 0nna

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 Σ an

is divergent.

If we find that , we know nothing

about the convergence or divergence of Σ an.

SERIES Note 3

lim 0nna

lim 0nna

Remember the warning in Note 2:

If , the series Σ an might converge or diverge.

SERIES

lim 0nna

Note 3

If Σ an and Σ bn are convergent series, then

so are the series Σ can (where c is a constant),

Σ (an + bn), and Σ (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

THEOREM 8 ii—PROOF

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

THEOREM 8 ii—PROOF

Hence, Σ (an + bn) is convergent, and

its sum is:

1

1 1

n nn

n nn n

a b s t

a b

THEOREM 8 ii—PROOF

Find the sum of the series

The series Σ 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 doesn’t

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