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Chapter 4. Series

4.1 Infinite series

4.1.1 Definition.

Given a sequence of numbers {an

}, an expression of

the form

a1+a2+a3+ +an+

is called an infinite series. The term an is the nth

term of the series.

For example,

1

2+

1

4+

1

8+

1

16+

1

32+

1

64+

is an infinite series whose nth term is 1

2n.

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The sequence{sn}defined by

s1 = a1

s2 = a1+a2

s3 = a1+a2+a3

...

sn = a1+a2+ +an=n

k=1ak

is called the sequence of partial sums of the series.

The numbersn is called thenth partial sum.

If the sequence of partial sums {sn} converges to a

limitL, then we say that the series is convergent and

that its sum is L. We write

n=1

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

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4 MA1505 Chapter 4. Sequences and Series

For this series, thenth partial sumsn is given by

sn = a+ar+ar2 + +arn1

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

Thussn rsn=a arn,

=sn=a1 rn

1 r, r= 1.

Ifr = 1, then clearly sn = na (or ), if

a= 0, and the series is divergent.

If|r|1, then|r|n , and the series diverges.

We summarize this as follow:

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5 MA1505 Chapter 4. Sequences and Series

4.1.3 Convergence of Geometric series

The geometric series

a+ar+ar2 + +arn1 +

witha = 0 converges to the sum a1 r

if|r|

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6 MA1505 Chapter 4. Sequences and Series

4.1.5 Some rules on series

If

an=A, and

bn=B, then

(1) Sum rule.

(an+bn) =A+B.

(2) Difference rule.

(an bn) =A B.

(3) Constant multiple rule.

(kan) =kA.

4.1.6 Ratio test

Let

anbe a series, and let

limn

an+1

an

=.

Then

(1) the series converges if 1.

(3) no conclusion if= 1.

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7 MA1505 Chapter 4. Sequences and Series

4.1.7 Example

(i)a1= 1,an+1= n

2n+ 1anand the series is

an= 1 +

1

3+

1 2

3 5+

1 2 3

3 5 7+ .

For this seriesan+1an

= n2n+ 1 12asn . Sothe ratio test implies the convergence of the series.

(ii)

(n!)2

(2n)!an+1an = (n+ 1)!(n+ 1)!(2n+ 2)! (2n)!n!n!

= (n+ 1)(n+ 1)

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

n+ 1

2(2n+ 1)

1

4.

So the given series is convergent by ratio test.

(iii) 3n

2n + 5an+1

an

= 3n+1

2n+1 + 5

2n + 5

3n = 3

1 + 5 2n

2 + 5 2n

3

2.

By ratio test, 3n

2n + 5is divergent.

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9 MA1505 Chapter 4. Sequences and Series

We state this as

1

1 x= 1 + x + x2 + + xn + , 1< x

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10 MA1505 Chapter 4. Sequences and Series

Case 1. The series cn(x a)n converges atx =aand diverges elsewhere.

Case 2. There is a positive numberhsuch that the

series converges for allxin the interval (a h, a + h)

but diverges for allx > a+h andx < a h.

The series may or may not converge at either of the

end pointsx=a handx=a+h.

Case 3. The series converges for everyx.

4.2.5 Radius of Convergence

The number h in case 2 of the previous section is

called the radius of convergenceof the series.

Note that we can describe all the points in the inter-

val in this case as |x a|< h. Hereais at the center

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11 MA1505 Chapter 4. Sequences and Series

of the interval.

If the power series converges for all x, we say that

the radius of convergence is infinite.

If it converges only at a, we say that the radius of

convergence is zero.

4.2.6 Example

(i) Find the radius of convergence of the power series

n=1

(1)n1x

n

n=x

x

2

2+

x

3

3 .

Solution. We apply ratio test to the series with nth

termun= (1)n1xn

n.un+1un = nn+ 1|x| |x| as n .

Therefore, the series converges for |x| < 1. It di-

verges if|x|>1.

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12 MA1505 Chapter 4. Sequences and Series

The radius of convergence is therefore equal to 1.

(ii) For the series

n=0

xn

n! = 1 +x+

x2

2!+

x3

3!+

un+1

un

=

xn+1

(n+ 1)!

n!

xn

= |x|

n+ 10 as n .

Therefore, the series converges for allx.

The radius of convergence is therefore equal to .

(iii) For the series

n=0n!xn = 1+x+2!x2+3!x3+ .

un+1un =

(n+ 1)!xn+1n!xn = (n+ 1)|x|

asn unlessx= 0.

Therefore, the series diverges for allxexceptx= 0.

The radius of convergence is zero.

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4.2.7 Differentiation and Integration of power

series

If

cn(x a)n has radius of convergenceh,

it defines a functionf:

f(x) =

n=0

cn(x a)n, a h < x < a+h.

(i) The function fhas derivatives of all orders in

(a h, a+h). The derivatives can be obtained by

differentiating the power series term-by-term:

f(x) =

n=1ncn(x a)

n1,

f(x) =

n=2

n(n 1)cn(x a)n2, . . . .

The differentiated series converges for a h < x