8/10/2019 NUS - MA1505 (2012) - Chapter 4
1/27
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.
8/10/2019 NUS - MA1505 (2012) - Chapter 4
2/27
2 MA1505 Chapter 4. Sequences and Series
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.
2
8/10/2019 NUS - MA1505 (2012) - Chapter 4
3/27
8/10/2019 NUS - MA1505 (2012) - Chapter 4
4/27
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:
4
8/10/2019 NUS - MA1505 (2012) - Chapter 4
5/27
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|
8/10/2019 NUS - MA1505 (2012) - Chapter 4
6/27
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.
6
8/10/2019 NUS - MA1505 (2012) - Chapter 4
7/27
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.
7
8/10/2019 NUS - MA1505 (2012) - Chapter 4
8/27
8/10/2019 NUS - MA1505 (2012) - Chapter 4
9/27
9 MA1505 Chapter 4. Sequences and Series
We state this as
1
1 x= 1 + x + x2 + + xn + , 1< x
8/10/2019 NUS - MA1505 (2012) - Chapter 4
10/27
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
10
8/10/2019 NUS - MA1505 (2012) - Chapter 4
11/27
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.
11
8/10/2019 NUS - MA1505 (2012) - Chapter 4
12/27
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.
12
8/10/2019 NUS - MA1505 (2012) - Chapter 4
13/27
13 MA1505 Chapter 4. Sequences and Series
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