02 Series and the best for the first website which is .pdf

One of the best for the day for you to get the same as soon varum i am not having laptop with you for the train and will send the link for this iq and I will send the details for your room no problem and I will be late by the way and then I can understand that

1. MA1000: Calculus
S Vijayakumar
Indian Institute of Information Technology,
Design & Manufacturing, Kancheepuram

2. Series
Consider a tiny frog which is initially at the point 0 of the number line. It makes successive
rightward jumps along the number line as follows.
It jumps 1 unit in step 1.
It jumps
1
2
units in step 2.
It jumps
1
4
units in step 3.
It jumps
1
8
units in step 4.
It jumps
1
16
units in step 5.
.
.
.
Thus the total distance travelled by the frog is
1+
1
2
+
1
4
+
1
8
+
1
16
+: : : :

3. The total distance travelled by the frog is
1+
1
2
+
1
4
+
1
8
+
1
16
+: : ::
Partial sum Partial sumValue
First: s1=1 21 1
Second: s2=1+
1
2
2
1
2
3
2
Third: s3=1+
1
2
+
1
4
2
1
4
7
4
.
.
.
.
.
.
.
.
.
.
.
.
nth: sn=1+
1
2
+
1
4
+: : :+
1
2
n1
2
1
2
n1
2
n
1
2
n1
The sequencefsngof partial sums converges to 2.
In this case, we say that the seriesconvergesto 2.
And say that thesumof the series is 2.

4. Denition (Series,nth Term, Partial Sum, Converges, Sum)
Given a sequencefang, an expression of the form
a1+a2+a3+: : :+an+: : :
is called aseries. The numberanis called thenth term of the series. The sequencefsng
dened by
s1=a1
s2=a1+a2
s3=a1+a2+a3
.
.
.
sn=a1+a2+a3+: : :+an
.
.
.
is called thesequence of partial sumsof the series and the numbersnis called thenth
partial sum.

5. DenitionContd.
If the sequencefsngof partial sums converges to a limitl, we say that the seriesconverges
and that itssumisl. In this case, we also write
a1+a2+a3+: : :+an+: : :=
1
X
n=1
an=l:
If the sequence of partial sums of the series does not converge, we say that the seriesdiverges.

6. Example: Geometric Series
Ageometric seriesis a series of the form
a+ar+ar
2
+: : :+ar
n1
+: : :=
1
X
n=1
ar
n1
:
Hereaandrare xed real numbers anda6=0. Thecommon ratiorcan be positive or
negative or even zero.
We will now prove that the geometric series converges for eachrwithjrj<1 and it diverges
otherwise.

7. Geometric Series
Case 1:r=1:Thenth partial sum of the geometric series is
sn=a+a(1) +a(1
2
) +: : :+a(1
n1
) =na:
So, the sequence of partial sums diverges to1or1depending on whethera>0 ora<0.
Hence, in this case, the series diverges.
Case 2:r=1:Thenth partial sum of the geometric series is
sn=a+a(1) +a(1) +a(1) +: : :+a((1)
n1
):
So, the sequence of partial sums diverges as it oscillates betweenaand 0. Hence, in this case
too, the series diverges.

8. Geometric Series
Case 3:jrj 6=1:
sn=a+ar+ar
2
+: : :+ar
n1
rsn=ar+ar
2
+: : :+ar
n1
+ar
n
snrsn=aar
n
sn(1r) =a(1r
n
)
sn=
a(1r
n
)
1r
(r6=1)
Thus ifjrj<1, thenr
n
!0 asn! 1and sosn!
a
1r
;
ifjrj>1, thenjr
n
j ! 1and the series diverges.

9. Geometric Series: Summary
Ifjrj<1, the geometric seriesa+ar+ar
2
+: : :+ar
n1
+: : :converges toa=(1r):
1
X
n=1
ar
n1
=
a
1r
;jrj<1:
Ifjrj 1, the series diverges.

10. Example
The geometric series witha=1=9 andr=1=3 is
1
9
+
1
27
+
1
81
+: : :=
1=9
1(1=3)
=
1
6
:
Homework:Find the sum of the geometric series
1
X
n=0
(1)
n
2
5(3
n
)
=
2
5

2
53
+
2
59
: : : :

11. Homework
1. ameters above a at surface. Each time the ball hits the surface
after falling a distanceh, it rebounds a distancerh, whereris a positive constant that is
less than 1. Find the total distance the ball travels up and down.
2. :232323: : :as the ratio of two integers.
3.
1
X
n=1
1
n(n+1)
=
1
12
+
1
23
+: : :+
1
n(n+1)
+: : : :
4.
1
X
n=1
n
2
diverges.
5.
1
X
n=1
n+1
n
diverges.

12. More Examples
Find the sum of the series
1
X
n=1
1
n(n+1)
=
1
12
+
1
23
+: : :+
1
n(n+1)
+: : : :
Solution:Here thenth term
1
n(n+1)
can be written as
1
n(n+1)
=
1
n

1
n+1
(n=1;2;3; : : :):
So, thenth partial sum can be written as a telescoping sum:
sn=
1
12
+
1
23
+: : :+
1
n(n+1)
=

1
1
2

+

1
2

1
3

+: : :+

1
n

1
n+1

=1
1
n+1
:
Thussn!1. Hence the sum of the series is 1.

13. Diverging Series
Show that the series
1
X
n=1
n
2
diverges.
Solution:Here thenth partial sum issn=
n(n+1)(2n+1)
6
.
Obviously,sn! 1asn! 1. Hence the series diverges.
Show that the series
1
X
n=1
n+1
n
diverges.
Solution:Here each term of the series is greater than 1. So, thenth partial sumsn>n. Thus
sn! 1asn! 1. Hence the series diverges.

14. An Important Theorem
Theorem
If
1
X
n=1
anconverges, thenan!0.
Proof.
ILet
1
X
n=1
an=l.
IThis means that the sequencefsngof partial sums converges tol.
IButan=snsn1andsn!landsn1!l.
IThusan=snsn1!ll=0.

15. Thenth Term Test for Divergence
Iflim
n!1
andoes not exist or is dierent from 0, then the series
1
X
n=1
andiverges.

16. Applying thenth Term Test for Divergence
1.
1
X
n=1
n
2
diverges becausen
2
! 1.
2.
1
X
n=1
n+1
n
diverges because
n+1
n
!16=0.
3.
1
X
n=1
(1)
n+1
diverges becauselim
n!1
(1)
n+1
does not exist.
4.
1
X
n=1
n
2n+5
diverges because
n
2n+5
!
1
2
6=0.

17. The Converse of the Theorem is not True
an!0 does not imply that
1
X
n=1
anconverges.
Example:For the series
1+
1
2
+
1
2
+
1
4
+
1
4
+
1
4
+
1
4
+: : :+
1
2
n
+
1
2
n
+: : :+
1
2
n
+: : : ;
thenth terman!0. But the series diverges as its partial sumssnincrease without bound.
Indeed,
s2>1;s4>2;s8>3;s16>4; : : : ;s2
n>n; : : : :

18. Note
We will often denote the series
1
X
n=1
ansimply as
X
an.

19. Combining Series
Theorem
Let
1
X
n=1
an=aand
1
X
n=1
bn=bbe convergent series. Then
1.
X
(an+bn) =a+b.
2.
X
(anbn) =ab.
3.
X
(kan) =kafor any numberk.

20. Proof (1):
ILetAn=a1+a2+a3+: : :+anandBn=b1+b2+b3+: : :+bn.
IThen the partial sums of the series
X
(an+bn)are
sn= (a1+b1) + (a2+b2) +: : :+ (an+bn)
= (a1+a2+a3+: : :+an) + (b1+b2+b3+: : :+bn)
=An+Bn:
IButAn!aandBn!b.
IHencesn=An+Bn!a+bby the Addition Rule for sequences.

21. Corollary
1.Every nonzero constant multiple of a divergent series diverges.
2.If
X
anconverges and
X
bndiverges, then
X
(an+bn)and
X
(anbn)both diverge.

22. Example
Find the sum of the series
1
X
n=1
3
n1
1
6
n1
:
Solution:
1
X
n=1
3
n1
1
6
n1
=
1
X
n=1

1
2
n1

1
6
n1

=
1
X
n=1
1
2
n1

1
X
n=1
1
6
n1
=
1
1(1=2)

1
1(1=6)
=2
6
5
=
4
5
:

23. Note
Addition or deletion of a nite number of terms does not aect the convergence or divergence
of a series. But in the case of convergent series, this may change the sum of the series.

24. An Important Series
Show that the series
1
X
n=0
1
n!
=1+
1
1!
+
1
2!
+
1
3!
+: : :+
1
n!
+: : :
converges.
Solution:Here thenth partial sum is
sn=1+
1
1!
+
1
2!
+
1
3!
+: : :+
1
n!
<1+1+
1
2
+
1
2
2
+: : :+
1
2
n1
<1+1+
1
2
+
1
2
2
+: : :+
1
2
n1
+: : :
=1+2
=3
The sequence of partial sums is bounded above by 3. It is also monotonically increasing.
Thus the series converges.

25. An Important Series
Denition
e=
1
X
n=0
1
n!
=1+
1
1!
+
1
2!
+
1
3!
+: : :+
1
n!
+: : : :
Note:e=2:7182: : :

26. Example: The Harmonic Series
The series
1
X
n=1
1
n
=1+
1
2
+
1
3
+: : :+
1
n
+: : :
is called theharmonic series.
The harmonic series is divergent:
1+
1
2
+

1
3
+
1
4

+

1
5
+
1
6
+
1
7
+
1
8

+

1
9
+
1
10
+: : :+
1
16

+: : :
>1+
1
2
+ +
1
2
+ +
1
2
+: : : :

27. Tests of Convergence for Series: The Integral Test
Theorem (The Integral Test)
Letfangbe a sequence of positive terms. Suppose thatan=f(n), wherefis a continuous,
positive, decreasing function ofxfor allnN(Na positive intger). Then the series
1
X
n=N
an
and the integral
Z
1
n=N
f(x)dxboth converge or both diverge.

28. Proof
We will prove the theorem forN=1. Letfbe a function withf(n) =anfor alln.
The rectangles in Figure (a) which have areasa1;a2; : : : ;ancollectively enclose an area
a1+a2+: : :+an. This is more than the area under the curvey=f(x)from 1 ton+1. That
is,
Z
n+1
1
f(x)dxa1+a2+: : :+an:
Note:

29. Proof:
The rectangles in Figure (b) having areasa2;a3; : : : ;ancollectively enclose an area
a2+a3+: : :+an. This is less than the area under the curvey=f(x)from 1 ton. That is,
a2+a3+: : :+an
Z
n
1
f(x)dx:
Addinga1to both sides, we get
a1+a2+a3+: : :+ana1+
Z
n
1
f(x)dx:

30. Combining the two inequalities gives
Z
n+1
1
f(x)dxa1+a2+: : :+ana1+
Z
n
1
f(x)dx:
The above inequalities hold for eachnand continue to hold asn! 1.
If
Z
1
1
f(x)dxis nite, the right-hand inequality implies that
P
anis nite.
If
Z
1
1
f(x)dxis innite, the left-hand inequality implies that
P
anis innite.
Hence the series and the integral are both nite or both innite.

31. Example: Thep-Series
The series
1
X
n=1
1
n
p
=1+
1
2
p
+
1
3
p
+: : :+
1
n
p
+: : :
is called thep-series (pany real constant).
Thep-series converges ifp>1 and diverges ifp1.

32. Example: Thep-Series
Show that thep-series
1
X
n=1
1
n
p
=1+
1
2
p
+
1
3
p
+: : :+
1
n
p
+: : :
converges ifp>1 and diverges ifp1.
Solution:Ifp>1, thenf(x) =
1
x
pis a positive decreasing function ofx(forx1). Now
Z
1
1
1
x
p
dx=
Z
1
1
x
p
dx
=

x
p+1
p+1
1
1
=
1
1p

1
x
p1

1
1
=
1
1p
[01] =
1
p1
:
So, the series converges by the integral test in this case.

33. If 0<p<1, then 1p>0 and
Z
1
1
1
x
p
dx=
Z
1
1
x
p
dx=

x
p+1
p+1
1
1
=
1
1p

x
1p

1
1
=1:
Ifp=1, we have the (divergent) harmonic series:
1+
1
2
+
1
3
+: : :+
1
n
+: : : :
Thus thep-series converges forp>1 and diverges for 0<p1.
Also thep-series diverges ifp0. (Why?)

34. Homework
1.
1
X
n=1
1
n
2
+1
converges using the integral test.
2.
1
X
n=1
1
n(logn)
p
converges if and only ifp>1 using the integral test.

35. Tests of Convergence of Series: The Comparison Test
Theorem (The Comparison Test)
Let
P
anbe a series of non-negative terms. Then
(a)
P
anconverges if there is a convergent series
P
cnwithancnfor allnN, for some
integerN.
(b)
P
andiverges if there is a divergent series of non-negative terms
P
dnwithandnfor all
nN, for some integerN.
Proof:In Part (a), the sequence of partial sums of
P
anis a monotonically increasing
sequence and is bounded above by
M=a1+a2+: : :+aN1+
1
X
n=N
cn:
So it converges.

36. In Part (b), the sequence of partial sums of
P
anis not bounded from above.
If they were, the partial sums for
P
dnwould be bounded above by
M
0
=d1+d2+: : :+dN1+
1
X
n=N
an
and
P
dnwould be converging!

37. Examples
1.
1
X
n=1
7
7n2
diverges because itsnth term
7
7n2
=
1
n
2
7
>
1
n
which is thenterm of the divergent harmonic series.
2.
1+
1
1!
+
1
2!
+
1
3!
+: : :+
1
n!
+: : :
converges because its terms are all positive and are less than or equal to the corresponding
terms of
1+1+
1
2
+
1
2
2
+: : :+
1
2
n1
+: : :
which is convergent.

38. Tests of Convergence of Series: The Limit Comparison Test
Theorem (The Limit Comparison Test)
Suppose thatan>0andbn>0for allnN(Nan integer).
1.Iflim
n!1
an
bn
=c>0, then
P
anand
P
bnboth converge or both diverge.
2.Iflim
n!1
an
bn
=0and
P
bnconverges, then
P
anconverges.
3.Iflim
n!1
an
bn
=1and
P
bndiverges, then
P
andiverges.

39. Proof (1):
Since
c
2
>0, there exists an integerNsuch that
nN)




an
bn
c




<
c
2
:
Thus, fornN,

c
2
<
an
bn
c<
c
2
;
c
2
<
an
bn
<
3c
2
;

c
2

bn<an<

3c
2

bn;
If
P
bnconverges, then
P
(3c=2)bnconverges and
P
anconverges by the direct Comparison
Test.
If
P
bndiverges, then
P
(c=2)bndiverges and
P
andiverges by the direct Comparison Test.

40. Example
Does the following series converge or diverge?
1
X
n=1
2n+1
(n+1)
2
=
1
X
n=1
2n+1
n
2
+2n+1
:
Solution:Letan=
2n+1
n
2
+2n+1
. Fornlarge, we expectanto behave like 2n=n
2
=2=n. So, we let
bn=1=n.
Since
1
X
n=1
bn=
1
X
n=1
1
n
diverges and
lim
n!1
an
bn
= lim
n!1
2n
2
+n
n
2
+2n+1
=2;
P
andiverges by Part 1 of the Limit Comparison Test.

41. Example
Does the following series converge or diverge?
1
X
n=2
1+nlnn
n
2
+5
:
Solution:Letan= (1+nlnn)=(n
2
+5). For largen,anwill behave likenlnn=n
2
= lnn=n,
which is greater than 1=nforn3. So, we takebn=1=n.
Since
1
X
n=1
bn=
1
X
n=1
1
n
diverges and
lim
n!1
an
bn
= lim
n!1
n+n
2
lnn
n
2
+5
= lim
n!1
1
n
+ lnn
1+
5
n
2
=1;
P
andiverges by Part 3 of the Limit Comparison Test.

42. Homework
Does
1
X
n=1
lnn
n
3=2
converge?

43. The Ratio Test
Theorem (The Ratio Test)
Let
P
anbe a series with positive terms and suppose that
lim
n!1
an+1
an
=:
Then
(a)the seriesconvergesif <1,
(b)the seriesdivergesif >1or is innite,
(c)the test is inconclusive if=1.

44. Proof
(a) <1.
Letrbe a number betweenand 1: <r<1.
Then the number=ris positive.
Since
lim
n!1
an+1
an
=;
there is an integerNsuch that
nN)




an+1
an





<
This implies that, fornN,
<
an+1
an
< or <
an+1
an
< +=r:

45. That is,
aN+1<raN;
aN+2<raN+1<r
2
aN;
aN+3<raN+2<r
3
aN;
.
.
.
aN+m<raN+m1<r
m
aN:
Consider the series
P
cn, wherecn=anforn=1;2; : : : ;NandcN+1=raN,
cN+2=r
2
aN; : : : ;cN+m=r
m
aN; : : :.
Nowancnfor allnand
1
X
n=1
cn=a1+a2+: : :+aN1+aN+raN+r
2
aN+: : :
=a1+a2+: : :+aN1+aN(1+r+r
2
+: : :)
The geometric series 1+r+r
2
+: : :converges asjrj<1. So
P
cnconverges.
Sinceancn,
P
analso converges.

46. (b)1< 1.
From some indexMon,
an+1
an
>1 and aM<aM+1<aM+2< : : : :
So, the terms of the series do not approach zero asnbecomes innite. Hence the series
diverges by thenth Term Test.

47. (c)=1.
Consider the following two series:
1
X
n=1
1
n
and
1
X
n=1
1
n
2
:
For
1
X
n=1
1
n
:
an+1
an
=
1=(n+1)
1=n
=
n
n+1
!1:
For
1
X
n=1
1
n
2
:
an+1
an
=
1=(n+1)
2
1=n
2
=

n
n+1

2
!1:
In both cases,=1. But the rst series diverges and the second converges.

48. Example
Investigate the convergence of the series
1
X
n=1
2
n
+5
3
n
:
Solution:Here
an+1
an
=
(2
n+1
+5)=3
n+1
(2
n
+5)=3
n
=
1
3

2
n+1
+5
2
n
+5
!
2
3
:
The series converges because here=2=3<1.

49. Example
Test the convergence of the series
1
X
n=1
2
n
n!
:
Solution:For the series
1
X
n=1
a
n
n!
,
an+1
an
=
a
n+1
=(n+1)!
a
n
=n!
=
a
n+1
!0:
The series converges because here=0<1.

50. Example
Test the convergence of the series
1
X
n=1
4
n
n!n!
(2n)!
:
Solution:Here
an+1
an
=
4
n+1
(n+1)!(n+1)!
(2n+2)!

(2n)!
4
n
n!n!
=
4(n+1)(n+1)
(2n+2)(2n+1)
=
2(n+1)
2n+1
!1:
Because the limit=1, we cannot decide whether the series converges from the ratio test.
But
an+1
an
=
2(n+1)
2n+1
>1 for alln:
Thusan+1>anfor alln.
So,a1<a2<a3< : : :.
Alsoa1=2. Thusandoes not converge to 0.
Hence the series diverges.

51. The Root Test
Theorem (The Root Test)
Let
P
anbe a series withan0fornN(Nan integer) and suppose that
lim
n!1
n
p
an=:
Then
(a)the series converges if <1,
(b)the series diverges if >1oris innite,
(c)the test is inconclusive if=1.

52. Examples
Which of the following series converges and which diverges?
(a)
1
X
n=1
n
2
2
n
(b)
1
X
n=1
2
n
n
2
(c)
1
X
n=1

1
1+n

n
(a)
1
X
n=1
n
2
2
n
converges because
n
r
n
2
2
n
=
(
n
p
n)
2
2
!
1
2
<1.
(b)
1
X
n=1
n
2
2
n
diverges because
n
r
2
n
n
2
=
2
(
n
p
n)
2
!
2
1
>1.
(c)
1
X
n=1

1
1+n

n
converges because
n
s

1
1+n

n
=
1
1+n
!0<1.

53. Example
Letan=

n=2
n
;nodd
1=2
n
;neven.
Does
P
anconverge?
Solution:Here
n
p
an=

n
p
n=2;nodd
1=2;neven.
Therefore,
1
2

n
p
an
n
p
n
2
:
We know that
n
p
n!1. So,lim
n!1
n
p
an=1=2 by the Sandwich Theorem.
Thus here the limit is <1. Hence the series converges by the Root Test.

54. Alternating Series
Denition
A series in which the terms are alternatively positive and negative is called analternating
series.

55. Examples
1
1
2
+
1
3

1
4
+: : :+
(1)
n+1
n
+: : :
2+1
1
2
+
1
4

1
8
+: : :+
(1)
n
4
2
n
+: : :
12+34+56+: : :+ (1)
n+1
n+: : :
The rst series, called thealternating harmonic series, converges.
The second series, a geometric series with common ratior=1=2, converges.
The third series diverges because thenth term does not approach zero.

56. Homework
Letfangbe a sequence such that the subsequencesfa2mgandfa2m+1gboth converge to the
same limitl. Then show thatan!l.

57. The Alternationg Series Test
Theorem (The Alternationg Series Test (Leibniz's Theorem))
The series
1
X
n=1
(1)
n+1
un=u1u2+u3u4+: : :
converges if all three of the following conditions are satised:
1.Theun's are all positive.
2.unun+1for allnN, for some integerN.
3.un!0.

58. Proof:
Let us assume thatN=1.
Ifnis an even integer, sayn=2m, then the sum of the rstnterms is
s2m= (u1u2) + (u3u4) +: : :+ (u2m1u2m)
=u1(u2u3)(u4u5): : :(u2m2u2m1)u2m
The rst equality shows that thes2mis the sum ofmnon-negative terms. Hences2m+2s2m.
The second equality shows thats2mu1.
So,fs2mgis monotonically increasing and bounded above. So, it converges, say
lim
n!1
s2m=l:
Alsou2m+1!0. Hence
s2m+1=s2m+u2m+1!l+0=l:

59. Thus we have thats2m!lands2m+1!l.
Hencesn!l. This means that the alternating series converges.

60. Example
The alternating harmonic series
1
X
n=1
(1)
n+1
n
=1
1
2
+
1
3

1
4
+: : :
satises the three requirements of the alternating series theorem withN=1. So, it converges.

61. Absolute and Conditional Convergence
Denition (Absolute Convergence)
A series
P
anconverges absolutelyif the corresponding series of absolute values
P
janj
converges.
Example:
The geometric series
1
1
2
+
1
4

1
8
+: : :
converges absolutely because the corresponding series of absolute values
1+
1
2
+
1
4
+
1
8
+: : :
converges.
In contrast, the alternating harmonic series does not converge absolutely: The corresponding
series of absolute values is the divergent harmonic series.

62. Denition (Conditional Convergence)
A series that converges but does not converge absolutely is said toconverge conditionally.
Example:The alternating harmonic series converges conditionally.

63. The Absolute Convergence Test
Theorem
If
1
X
n=1
janjconverges, then
1
X
n=1
anconverges.

64. Proof:
For eachn,
janj an janjso 0an+janj 2janj:
If
1
X
n=1
janjconverges, then
1
X
n=1
2janjconverges.
So, by the Comparison Test,
1
X
n=1
(an+janj)converges.
Butan= (an+janj) janj. So,
1
X
n=1
ancan be expressed as the dierence of two convergent
series:
1
X
n=1
an=
1
X
n=1
(an+janj janj) =
1
X
n=1
(an+janj)
1
X
n=1
janj:
Therefore,
1
X
n=1
anconverges.

65. Examples
(a)
1
X
n=1
(1)
n+1
n
3
=1
1
8
+
1
27

1
64
+: : :, the corresponding series of absolute values is
the series
1
X
n=1
1
n
3
=1+
1
8
+
1
27
+
1
64
+: : : :
The latter converges. Thus the original series converges absolutely. Hence it converges.
(b)
1
X
n=1
sinn
n
2
=
sin1
1
+
sin2
4
+
sin3
9
+: : :, the corresponding series of absolute values is
1
X
n=1




sinn
n
2




=




sin1
1




+




sin2
4




+




sin3
9




+: : : :
Sincejsinnj 1, the latter converges by comparison with
1
X
n=1
1
n
2
.
Thus the original series converges absolutely. Hence it converges.