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Math 55 chapter 18
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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.
Math 55 Series 2/ 20
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?
Math 55 Series 3/ 20
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.
Math 55 Series 4/ 20
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.
Math 55 Series 5/ 20
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.
Math 55 Series 6/ 20
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
Math 55 Series 7/ 20
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.
Math 55 Series 9/ 20
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.
Math 55 Series 11/ 20
Geometric Series
Example
Determine if the series
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.
Math 55 Series 12/ 20
Geometric Series
Example
Determine if the series
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.
Math 55 Series 14/ 20
Properties of infinite Series
Example
Determine if the series
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.
Math 55 Series 15/ 20
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 .
Math 55 Series 16/ 20
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.
Math 55 Series 17/ 20
Divergence Test
Example
Determine if the series
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.
Math 55 Series 18/ 20
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)
Math 55 Series 19/ 20
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/
Math 55 Series 20/ 20