18 series - handout

Series

Math 55 - Elementary Analysis III

Institute of MathematicsUniversity of the Philippines

Diliman

Math 55 Series 1/ 20

Recall

A sequence {an} = {a1, a2, . . .} is convergent if

limn an = L

exists for some real number L. Otherwise, it is divergent.

In many applications, the sum of the terms

a1 + a2 + . . .+ an + . . .

of a sequence is also important.


Series

Definition

An infinite series, or simply a series, is an expression of theform

n=1

an = a1 + a2 + . . .+ an + . . . .

The numbers a1, a2, . . . are called the terms of the series.

A Naural Question: Does this sum always exist?


Series

Consider the series

n=1

n = 1 + 2 + 3 + 4 + 5 + . . ..

Clearly as n increases, the sum becomes infinitely large.

Take the series 12 +14 +

18 +

116 + . . .+

12n + . . . and consider the

sum of the first k terms:

1

2+

1

4= 0.75

1

2+

1

4+

1

8= 0.875

1

2+

1

4+

1

8+

1

16= 0.9375

1

2+

1

4+

1

8+

1

16+

1

32= 0.96875

1

2+

1

4+

1

8+

1

16+

1

32+

1

64= 0.984375

1

2+

1

4+

1

8+

1

16+

1

32+

1

64+

1

128= 0.9921875

it seems that the series approaches 1.


Partial Sums

When does a series have a sum?

Let Sk denote the sum of the first k terms of the seriesn=1

an,

i.e.,

S1 = a1

S2 = a1 + a2

S3 = a1 + a2 + a3...

Sk = a1 + a2 + a3 + + ak =k

n=1

an

We call Sk the kth partial sum of the series and {Sk} the

sequence of partial sums.


Partial Sums

For

n=1

1

2n, with sum of the first n terms:

S1 =1

2= 0.5

S2 =2

n=1

1

2n=

1

2+

1

4= 0.75

S3 =3

n=1

1

2n=

1

2+

1

4+

1

8= 0.875

S4 =4

n=1

1

2n=

1

2+

1

4+

1

8+

1

16= 0.9375

S5 =5

n=1

1

2n=

1

2+

1

4+

1

8+

1

16+

1

32= 0.96875

S6 =6

n=1

1

2n=

1

2+

1

4+

1

8+

1

16+

1

32+

1

64= 0.984375

the sequence {Sk} = {0.5, 0.75, 0.875, 0.9375, 0.96875, 0.984375, . . .} ofpartial sums of the series seems to converge to 1.


Convergent/Divergent Series

Definition

Letn=1

an be a series and denote by {Sk} its sequence of partialsums. If {Sk} converges to S, then the series is said to beconvergent and its sum is equal to S; and write

n=1

an = S.

Otherwise, the series is said to be divergent.

In other words, a seriesn=1

an is convergent if its sequence of

partial sums {Sk} is convergent, orn=1

an = limk

Sk = limk

kn=1

an


Convergent/Divergent Series

Example

Determine if the series

n=1

1

n(n+ 1)converges or diverges.

Solution. Consider the kth partial sum Sk =k

n=1

1

n(n+ 1).

By partial fraction decomposition, note that1

n(n+ 1)=

1

n 1n+ 1

.

Sk =k

n=1

1

n(n+ 1)=

kn=1

(1

n 1n+ 1

)=

(1 1

2

)+

(1

2 1

3

)+

(1

3 1

4

)+ +

(1

k 1k + 1

)=

(1

1

2

)+

(1

2

1

3

)+

(1

314

)+ +

(1

k 1k + 1

)= 1 1

k + 1

So Sk 1 and therefore the series coverges,n=1

1

n(n+ 1)= 1.Math 55 Series 8/ 20

Harmonic Series

Example

Show that the harmonic series

n=1

1

nis divergent.

Solution. We consider the following partial sums:

S2 = 1 +1

2=

3

2

S4 =

(1 +

1

2

)+

(1

3+

1

4

)>

3

2+ 2

(1

4

)=

4

2

S8 =

(1 +

1

2+

1

3+

1

4

)+

(1

5+

1

6+

1

7+

1

8

)>

4

2+ 4

(1

8

)=

5

2

S16 =

(1 +

1

2+ + 1

8

)+

(1

9+ + 1

16

)>

5

2+ 8

(1

16

)=

6

2

S2n >n+ 2

2.

Sincen+ 2

2, the harmonic series

n=1

1

ndiverges.


Geometric Series

Theorem

The geometric seriesn=1

arn1 = a+ ar + ar2 + . . ., where a

and r are nonzero constants, is

(i) convergent if |r| < 1 andn=1

arn1 =a

1 r .(ii) divergent if |r| 1.

Proof. If r = 1, the geometric series becomesn=1

a.

The nth partial sum is Sn =n

i=1

a = na. Since

limnSn = limnna =

{+ if a > 0 if a < 0

the series diverges.Math 55 Series 10/ 20

Geometric Series

Proof.(contd) If r 6= 1,

Sn = a+ ar + ar2 + ar3 + . . .+ arn1

rSn = ar + ar2 + ar3 + . . .+ arn1 + arn

Sn rSn = a arn(1 r)Sn = a(1 rn)

Sn =a(1 rn)

1 r

Now, limnSn = limn

a(1 rn)1 r =

{ a1 r if |r| < 1dne if |r| > 1

and we get the

desired result.


Geometric Series

Example


n=1

pin1

3nconverges. If it does, find the

sum.

Solution. n=1

pin1

3n=

n=1

1

3

(pi3

)n1This is a geometric series with a = 13 and r =

pi3 .

Since |r| 1, the series diverges.


Geometric Series

Example


k=0

(2)k3k1

converges. If it does, find the

sum.

Solution.

k=0

(2)k3k1

=

k=0

(2)k313k

=

k=0

3

(2

3

)k

We recognize this as a geometric sequence with a = 3 andr = 23 . Since |r| < 1, the series converges and

k=0

(2)k3k1

=3

1 (23) = 95Math 55 Series 13/ 20

Properties of Infinite Series

1 Ifn=1

an andn=1

bn are convergent, thenn=1

(an bn) is

convergent andn=1

(an bn) =n=1

an n=1

bn.

2 If c 6= 0 andn=1

an is convergent, thenn=1

can is also

convergent andn=1

can = cn=1

an.

3 If c 6= 0 andn=1

an is divergent, thenn=1

can is also

divergent.

4 For any positive integer k,n=1

an = a1 + a2 + a3 + . . . and

n=k

an = ak + ak+1 + ak+2 + . . . are both convergent or

both divergent.


Properties of infinite Series

Example


n=1

(1

2n1+

2

n(n+ 1)

)converges or

diverges.

Solution.

n=1

1

2n1is a geometric series with r = 12 and

therefore, it is convergent.

We have seen earlier that

n=1

2

n(n+ 1)converges.

Therefore,

n=1

(1

2n1+

2

n(n+ 1)

)=

n=1

1

2n1+

n=1

2

n(n+ 1)converges.


A Property of Convergent Series

Theorem

Ifn=1

an is convergent, then limn an = 0.

Proof. For a seriesn=1

an with nth partial sum Sn, note that

an = Sn Sn1Ifn=1

an is convergent, then there exists a number S such that

limnSn = S. Hence,

limn an = limn (Sn Sn1)

= limnSn limnSn1

= S S= 0 .


Divergence Test

Corollary (Divergence Test)

If limn an 6= 0, then the series

n=1

an is divergent.

This theorem guarantees divergence of series if the terms do notapproach 0.

However, If limn an = 0, the series

n=1

an may or may not

converge. To illustrate, consider the series:n=1

1

nwhich is divergent;

n=1

1

n(n+ 1)which is convergent.


Divergence Test

Example


n=1

en

n2converges or diverges.

Solution.

limn an = limn

en

n2

= limn

en

2n

= limn

en

2=

Hence limn an 6= 0 and so the series is divergent by the

Divergence Test.


Exercises

Determine if the series converges or diverges.

1

n=1

ln

(1 +

1

n

)2

k=2

k2

k2 1

3

i=1

i

1 + ln i

4

n=1

1 + 3n

2n

5

n=1

2

n(n+ 2)


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/


18 series - handout

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