3

Sequences
1. INTRODUCTION

In the present chapter we propose to study a special class of func

rions, namely sequences.
ABefinition 1-1. Asequence in aset Sis afunction, whose do-
mainisthe set N of natural numbers and whose range is a subset of S.
Asequence whose range is a subset of R is called a real sequence.
Since we are going to deal with real sequences only, therefore we
chall use the term sequence to denote a real sequence.
From the above definition, it is obvious that a sequence assigns to
each naturalnumber n, a unique real number. It is usual to denote by a,
the real number assigned to the natural number n by a sequence a. This
notationdiffers from that for functions in general. For, a sequence a being
afunction from Nto R, the value of this function at n should be denoted
by a(n). We shall, however, denote it by a,. Asequence a is thus the set
(n, a,) : ne N}. Since the domain of all sequences is the same set,
namely N,therefore, a sequence is determined completely if we know a,
for each n. In view of this, it is usual to write a sequence as <a,> or as
<a, a, az ...>. We must, however, be careful to distinguish between a
sequence <a,> and its range {a,:ne N).
The numbers a,, a,, ... are called the terms (or elements) of the
sequence. Thus, a, is the first term, a, isthe second term, ... a, is the nth
term of the sequence <a,>.
Asequence can be described in several different ways.
We may list in order the first few elemnents of a sequence, till the
rule for writing downdifferent elements becomes evident. For exanple,
<I, 4,9, 16, 25, ..> is the sequence whose nth term is n.
Another method of describing a sequence is to give a formula for its
ih term. For example, the sequence <l, 4, 9, 16, 25, ...> can also be
written asS<l, 4,.. n,..> or as <u':ne N> or simply as <u>.
Another convenient nethod of describing asequence is to specify a,
and arule for obtaining 4, +I (for eachn2Din terms of dË, .., a Such
a definition is called arecursive definition. As an example of adefinition
by recursion, let
a, = l, a,+|=3a,, for alln > I.
 who anabor inteppr
number thisthan
numbers, madeand be we each the
less definition positive must If lie thesmaller
aj
sequence bythe
,of
and
terms Ato positive where be [- finite. sequence in in
above
convergeThe -0l=2", made course, ultimately
real specified can between relation[,l+ The limit'
in 2m. <a,>, 1/e)/log
(log
2. aboveandl some a the
be
define postive
a a, a 2m. except - [. [. 'a
n22. determine eUSIS can |/ the Of uponll-,l+[
re-state phrase limit'.
completely to all n sequence
la, -0l betweena, for above interval, lies
the lie n
said for have, whenever all of to depend
relationsany ). there in must
is
Efor
zero, a, >m'
the take, terms it
converge can toliff the 'ihe
be «, cantherrtore,SEQUENCES a, we
open
Sequences anda, 0. -ll< the > Ithat
CONVERGENT
2. that we thatphrase
+ considern allndifferencembols, may many willoutside converges
these =(a, we 2-1.
>
¬
sequence
to
if find
converges
i.e., <l+[
is the
cannot
discussion, shows
n, anv la, i.e., -!1<[for onwards, notewe in [,l+e the
that let
a, sequence. we sy lie
infinitely terms
number giventhat to >0 by
theorem
is. us l/[ the In l-¬<a, must sequence
seen anotherexample.
a,. <a,> let sequence
choose.
onwards.
of stage onwards. point
& - of
sequence
above replaced
if such "> Whatever ] number
natural l, example. 'la,value which
sequence outside
asily sequence numberthe important the following
. [). if mav phrase some the around
interval
I'. be
- MMefinition of 0) stage absolute stage 'a
of that
be each on limit Thisless for lying
the can
canis a fdependingan E> thanEwe
fromsome >0.the[ E[. view
the towards) some
As For a As n. positiveThe that The[ of one |l-¬l+ larger saying
terms In by
Thedefinition
It term andrelation. called all the c all
from thanfrom terms
even
nth d, for for of the
let m,
 Sincethebecause
one distinct ||-e.l+
intervals usually
contradiction
orexpressed (inlarge),= posi a con
a, be
than ultimately
both =
l' o lima to neither
0. a, + (however exists said
two> in impossible is lim
limit. to 2m.symbols,
more e 1,
t0 ,therefore, diverge it
Theone
limitor n
all
there
is
if
to converge o ; o0: oscillate
k for (in oo
converge thanthe o'
common. -
lie is
itmust to number smnall),or +
to - 1.
this moreto n said k> o to wherex> :
<u,> converges
las
o diverges diverges oscillates
But a, to (however
2m.+ 1. to
cannot Since
/ in to is positive
thatdiverge either > said ...>;
’ <a,>
Real
Analysis sequenceand
l',
therefore, converge
el. real
number x . > ; OSCILLATORY
|E- .>+
'a,
such n sequenceswhere ...>;
sequences ...>, SEQUENCES
is sequences
2,
sequence <a,>=lor SEQUENCES
DIVERGENT
3. sequence
each to k for to
all
-3", seauence 1Y'
E, m, numnberdiverges ..>; 2n, -x",
x..>, (-
apossible, -
let
If |/ no
and
cannot
sequence
a,
limn’oo
to
if
said k
integer
is
<
a,
2n...>
;
following ,-33
-
3", following
...,
..
..
following
2,
A = A negative r.x....6, A
both havesequence o) <a,>
positive thatwhich 3,..
- 3, 2,
2-I. Let
Converges
to
intervals
a writing
that ’.
3-1.+ sequence.
such Illustrations 4, 4-1.
verges
nor
diverges.
K -
Hustrations
Theorem I'. 'a,
simply Definition
=
a, sequence sequence the6, <3,3',the- - (c)
4, 2, <-3,
<-x,
- VDefinition
2,
the2,
eachm
Proof.and
numbers
/;
thefact by lim a
exists integer
of <2, <x. of <- of <-
twOthatThe i Each
divergent Each Each
lmit. Svmbols, to (b) (c) (a)
<a,> thesechows A if
there A (a) (a) (b)
90
oo) tive
- 4.
 seen
from
definitions h,+...
+
h, >0,
positive h,
the n-1 +1.
Then
since
then 2
two together.
above n, <E,
provided
m, all when
n2, n2.
when
be integers n(n
-1) m, m.
the q 1+h,. for 2
and oscillate h,y.
+
n=(l 2 2 2
=1. than I<[whenevern
n
..> h 1|<Ewhenever
Sequences 2n andp positiveVn Vn = +
n=(1
h,y', =1+
nh,
+ -1)
n(n , Vn-1 greater
..-1n. n-1 Vn-1
1. or
sequences lim 2 2 2
-2n casily diverge a,= 2 l.
all that
Prove write theorem. lh,
l< lh,
l< integer
providedn>+
1. =
... Then
V /n
3,4. be for converge. Th,
-
can twob,m us given. positive lim
Vn
2.- Let binomial
It be m= 1.Example |
1, be a
<- <1. <h,> <h,>
() Remark. Solution. [>0 be
.
a, the m
(b) andthatand
<a,> By Let Let
<a,>such that
so so
that Hence
L.e.,
Or Or
 ! that n, that that
corresponding
..i) ...(1) any
= such such
a, such
for
by that integer 0
h+..
h', all
n.for 0 > m
defined z &, integer the
such a, choose
positive that m,
multiplying
h>0 positivem.
<a,> /h. such ’0. can
-l| alln n.
nh
1+ m. 2
n(n
-l) 1 +||I,
an
sequence n. - a > a, we =h<l.
sequence ||+¬, whenever
n
exists 2 all (1/[be nwhenever a for 2n and have
Real
Analysis 1+h for (1+h)" m
thenl|<1,therefore,
is all l
+ >
if
there 1,
there nh nh nprovided zero. l|<1, +l - we
for ...,n
the |rl= therefore.
|+ >]+ (rl"=
l=ll=
la,
that a therefore,
Eo <h,
inequalities,
then I<eto be if| E
that lr||
lr|<1,
=
hy"
find
<a,>
converges
< < 1,
+
Show E, we la, that m
(1+ given.
< (ii), Let PrOve Since 1, m,
0, nh) then a,, by resulting
if > ’
2. ExampleSolution.
¿eroIf hsince be + and
[>01/( (i) 1)/h,
3.
<u,>
Example
’1. Solution. ,,
+1
neplacing
converges
to
Also. ThenFrom(1/[- Hence Since the
Let of
Now sides
92 m> and i.e.,
 dn t < h,
m

ln+ 2
Or <h
m
(n+| , -1

Or la, I< h" for all n2 m.

Also, since ) <h<l, therefore, h" ’0 and

E>0, there is a positive integer psuch that consequently, gven
Ih"|<h" e/ l a,, l, for all n2p.
From () and (ii), we find that
la, l<[, for all n2 max. {m, p).
Hence a, ’0.
5. BOUNDED SEQUENCES
Definitions 5-1. A sequence <a,> is said to be bounded (re.
spectively bounded above, bounded below) if there exists a real number
ksuch that Ia, Isk (respectively a, k, a, > k) for all n.
From the above definitions it is clear that a sequence is
there exist real numbers k and Ksuch that ksa,s K, for all n.bounded
if
It is not necessary that a sequence be bounded above or
bounded
below.
1llustrations
1. The 1
sequence <1, 2 3
-,..> is bounded because ||/n|<1 for all
ne N,
2. The sequence <u,>, where a, = (- 1", is bounded because l a, ls
1for all ne N. (In fact, I a, |= 1for all n e
N.)
3. The sequence defined by a, = n is bounded below but is not
bounded above.
4. The sequence defined by a, = -n is bounded above but is
bounded below. not
5. The sequence <(-1)" n> is neither bounded above nor bounded
blow.
Theorem 5-1. Every
Every convergent sequence is bounded.
Proof. Let <a,> be a sequence which converges to . Let us
choose E=1.Then there exists a positive integer msuch that
 Real Analysis
94

la, -1|<1, for all n m,

|-l<a, <I+ 1, for all n >m.
i.C.
k= min. 41, ag. am-,-1),
If
K= max. aj, az . am -|:+1).
then kSa, S K for alln.
Hence <a,> isbounded.
Remark. The converse of the above theorem does not hold. That
is,. abounded sequence may not converge. For example, the sequence
cl.-1.1. ..> is bounded, but is not convergent.
6. SOME IMPORTANT LIMIT THEOREMS
 herel sequence such >
integer c, therefore, n
=| if all
that m, for
, the integer
positive showing
im bounded,
be
Since |<Ewhenevern <l+¬. n. <S,> l. positive +
2q. a 2r. =
is S, to
number. there n r}. n if lim amounts is
sequence
Ewhenever
q. Sc, and
-|<ewhenever
Sequences theretore, {p. 1-E<h,
<l+E, a n.
that Sb, =l then is all
max. now there
positive a, a,)ln,
for
convergent
such |-E<a, lim b,
=l. I<h
-l< = lim +C,) theorem Then,
q. 1.
= m have t -I. ’0.
c,
any integerla, c, If +... l
tbe
lc, we lb. a, t... tc,) given.m. that
every
lim m, 6-11. a, = 0, the
>
c,
Let positive since + f ,
(a, t of t..be n such
all since
n Theorem proof 0
Proof. if = Proof. ( (t,
[> for h,
Also, Then. S, l+ number
Also,
a Let The Let
is as =
there so whence
that Hence fined S, <2 Now
that then and then lc, a
is
 be limdeducen.all
can
as
Delhi on
theorem for
nottrue, can 0
p}. >
that we a,
then
(m,
’0 theoremis
consequently,
and that
firsttheorem,
m, Ne so
>
max.
-’l. evens odd,
Cauchy's suc
n 21nh <-
whenever
n>p. He
all m. n> t...ta,) is
for n> than: all above is n this
Real
Analysis ifn if
-m), all greater yields for <a,>, as of
known
for <[ the 0. conseque
(n
, integer then)| a,
+
n
of (-1)".
=
a,
sequence += n-
limits
convergent.
2n 2 +c, +C, (a,
converse a,)
t ’0. usually
2
mh inequality
t... t...
on
mh
mh + positive S,=
the For t...
sequence,
this B3+..ta, theor
t, tC, Theconsidering a, Also,is Importa
not
<a,>iS
ag n 6-11
above Kemark.
+t
Iheoremsecon
ea ( C (a,
Ifp The Thus. dl Cauch
100 by
seen where
and
 that such integerm positive
fini- bounded. by=1, Taking Sequence. Cauchy exiast must there 7-1, tion
sequenceis Cauchy Every abe <u,> Let Proof.
rheorem
l<ewhenever
>0. p la,p-
a,,
7-1.
0given sequence
&> that such integerm positive
if Cauchy exIStaS there
following equi
the valtoent
abeto said is sequence
<a, > A
choiof
m. ce alteringthe Defi
is n
7-I itio n
<E 4, -a, "lI<Ewhenever onlamounts
to y change m. T whenevern>
his 2.
bym', 2n
may detfinition
we8-1, inthat seen easily
a, la-, phrase replace
the
becan RemarkS. I 1.
|< m - nwhi|ch for integer n>m. al for e
we1,Taki
[=ng what ever m positive no there
is that
be. may m >l, 2m >m) +(nm) -(n find
then m, >nsequence.if For, Cauchy = m-
sequence. Cauchy sequence <n>is
no
at sequence The 2.
ais given
m.that<e,
ow; This whenever the that
>n find we thangreater be
tto
akinm
g then given,
by e>0
be If nm
then m, >
r,isequence.
t. Cauchy ..>is a 1,, sequence
2 < The 1.
1 1
Mustrations X
l<Ewhenever
Cauchy th¡t m.> n
such integerm positive existsa there
a,-la,
0. gi>ven¬ quenceif
a beto said <a,is>sequence Definition
A7-1.
wdethetcrhe\eie:whinirtmecrhiionninsequence
g to the elements
of involves
the only
apply
it. towish
itthat fact the
imporCrile tant this utility
of and power
for Convergence
Criterion,
The not.convergesor sequence
Cauchy knownas rion,
a
an
establish propose
to section,
we present the In
SEQUENES 7.CAUCHY
lim that prove then L>Iand
0, >k If 0.

lim thennumber,
n real ny be tifthat Prove 29.

Srqences
 ...(2)
integers
Inn. positive that such
mainfinitely
ny integer
mk>
-ll<el3
for
positive finad a,thcaatnl so
weS,
pOiof
nts
particular,
manyinfinitely contain neighbourhood
nmust /of Every l.say poimt, limit
Bolzano-Weierstrass
ahtheorem,
as S Infinite.
By beSlerFirst,
finite. infinite
or accordjng
is Sas arise Cases diferent Two
bounded
set. therefore
ais S bounded, <a,is>Sequence the Since
N.:ne {u, S=
...() whenever
m2 n e/3 I< Ta,-a,
Lei
that such integerm positive exists| a theretherefore.
uence. Cauchy a<u,
is> Since number. positive any be E Let
<u,istherefore,
> bounded, sequence bounded.
sequence. Cauchy is Cauchy every Sice
a<a,is> thassume
at now Conversely,
uslet
sequence. Cauchy a<a,is> Hence
m.
whenever
2n /2, [<[I2+
-ll+la,,
-I, sla,
)I, (a,,-l=l(a,-
-) a,, Ia,- Now.
-1|<e/2. la,m
particular., In
whenever
m. 2 n -l|<[2 la,
that such integerm positive exista must there e>0. Given
uence. convergent ais<a,> thatassume
/.be
limit its Let
uslet First, Proof.
sequence. Cauchy a isii onlfy and comverges
quence Criterion),
A Convergence (Cauchy Theorem
7-2.
if

Cauchy sequence.
nota bounded
isbut is1)" (- =a,defined
by scquence
exFor true. nottheorem thamnle.
e
is above tconverse
he of ARemark.
The
bounded. <a,>
is Hence
n.al for kSa,s k Then

Let
whenever
m. 2 n I +a, I<a,<
<lwhenever
m,n2 a, Ia,
Analysis Real 104
 = such convergt.
a, 2
’1. n, -I'|<[by [ m
l), Taking
integer
a, indices
for defined
- -I1, consequently 1 cannot
(a, la, converges.
la, Sbe
many that positive
have + have al+ <a,> m, m. 1 2m sequence
+
a) a,l 2
we - Then such we
finite,
infinitely n n t...t ->m.
(a,, - [/3. and m (1) - 1. sequence
does
not
a
<a,> exists
converge. all all 2m
(1) la,, I+tla,
’ l<for for the
Sequences
form a,) + a,,l+ m > from that +2:
m
+ 2 k' a, that there Hence
e/3 n integer thatthe
-Ilsla,-a,, 2 see
therefore--1|=l(a, therefore
l<e/3. -la, +e/3 -|<ewhenever la-l'l<[l3
a,,la-have showing
l<e/3. that assume criterion,
a,, n| we +1
m
+
+2
m contradiction.
number.
positive we - 2m,
S < Show la,
m, a,, m,
k'>
(3)
m, +..+ us
Let convergence +2
=
n
m taking
-+
+1
m
> - la, reala andla, AExample
k Ia,
since find since n 4. a
nm.
whenever some (2) whenever Solution.
a, let can on have
lThusNext, =1++
3 + +]
m
Also, Now, is We Also, From But
Cauchy we
' 2
where Thus
l.e.,
 Real Analysis
108
8. MONOTONIC SEQUENCES
Mpefinitions 8-1. Asequence <a,> is said to be monotonically
increasing(or non-decreasing) ifa, . a, for all n.
Asequence <a,> is said to be monotonically decreasing (or non
increasing) f a, +|S a,for all n.
Asequence is said to be monotonic if it is either monotonicallv
increasing or monotonically decreasing.
Jllustrations
4. Each of the following sequences is monotonically increasing
(a) <l,2, 3, ... n, ...>
(b) <2, 2, 4, 4, 6, 6, ...>;

(c) <-I, 2 3 4
2. Each of the following sequences is monotonically decreasing:
(a) <I,
234

(b) <l, I,
3'3's'5
(c) <-2, - 4, - 6, 8, .>.
3. None of the following sequences is monotonic
(a) <0, I,0, I,0, 1. ...>:
(b) <- 2,2, - 4, 4, - 6, 6, ...>:

(c) <l,
3'5 7
important criterion for
Ine tollowing theorem gives a simple but
COnvergence of sequences Theorem) Every
Theorem 8-1. (Monotone Convergence
bounded monotonically increasing sequence converges.
Proof. Let <a,> be a bounded monotonically increasing sequence.
Let Sbe the {a, :n N). Then Sis a non-empty bounded set. By the
set number l= sup S. We
complshow property of real numbers, there is a
etenessthat <u,> converges tol.
shall
 Sequeces

Let ebe any positive real number. Then /-e<I, so that -

an upper bound of S. Therefore, a, >l- Efor some positivein

Since <a,> is monotonically increasing, theretore, integer m
a, 2 ,, >l-[for all n2 m.
Also, since /is the supremum ofS, therefore,
a, SI<l+[ for all n.
From () and (ii), we have
l-E<a, <l+ efor all n 2 m,
i.e. la, -1|<[ for all n2 m.
Hence <a,> converges to I.
Corollaries. 1. Every bounded monotonically
quence converges.
decreasing
Proof. Let <a,> be a bounded monotonically decreasing sequenr.
and let b, = - a,, for each positive integer n. Then <b,> is a bounde
monotonically increasing sequence and therefore, it converges. If lim
=l, then
lim a, = lim (-b,) =-lim b, =-1.
l ’ oo n’oo

2. Every bounded monotonic sequence converges.

As an application of the monotone convergence theorem, we prove
below the Cantor's theorem on nested intervals which ranks in impor
tance with Bolzano-Weierstrass theorem.
 n=]

Pxample 6. Show that the sequence <a,> defined by

1 1 1
+ t... t
n +| n +2 n + n

converges.

1 1
Solution an +| -a, = n+2 +
n+3
+...+ -
2n +2
-||5
1 1
-+ -+...+
n+1 n+2 2n
1
2n+1 2n +2 n+1

>0for all n.
2n +1 2n +2

Therefore, he sequence <a,> is monotonically increasing

Also,
1
la,l= + +...+
n +| n+2 n+n n

i.e., Ia,l<Ifor all n.

 fore, i.e., thannow rion, conyerges.
fore, it 12
Siitnce >
<a, From Also, Therefore. show
or Eorn>
2, ThereSolution. a,
must
is
Since equal
that I.
mpl= e Therefore.
JiaSequence Since
(i) <a,>
<a,>
converge. bounded.
>0
a,
and to fore, it
a,
n!=1.2.3
tw0, isthe a,+|- =Prove
<a,> 7. isthe
is for (ii), 1+
a n! bounded
sequence a
sequence
bounded,
bounded 0<a, all a, =l<3. we a <3, a,
1+
= 1 so
=l+ s1+1+
n, < find that a,
therefore, 3, 2"-1 1 n n as
..
= 2! thatby Real
< for <a,> <a,>
monotonically 3, a l that whenever 2!
1! !22" has well. n! monotonically
+ rela-
the
for whenever Analysis
n. 20-1
2 1. 1.1 -n
is >0 is
alln, from monotonically for bounded.
n
t...t 3! wheneverfactors 1 1)(n
!-
2. all
(ii), +...+
>2, n n. increasing
increasing each
we 27-2 so n> (n
find (n-1)! that increasing.
2. defined
2
of 2),
sequence, that greater
which is converges. sequence,
We
there ...(ii) ...() shall there
..()
 isThe
fact value
we we
Itsplaces, lies theorem,1
callede. The
number.
ten
To
and
exists binomial(n
-2) -)-2).1-)
24-T
t..+ transcendental n!
example places.e=2.7182818284.
nn-1) 1,1.+
t...t
1!2! +
3!1!2! the 3!
Sequences
above
l
e=+ t t
decimal lim|
1+ By 1, 2!
3! 22
a
=|1+-. +
+2
be
n’o 2
the that 1-1)
n(n +1+
<I
2<a, <l+|+
lim
e= to of n
in writing thousands
eisknown
limit Prove 2! =|+
that
a, find
The Let +n-+
by
indicated to
number 8. we
Remark. calculated Example Solution.
I
(i),
= From
is a,
been and
3.
 ...(iü)
n+|
>-)-
(1-).
-above)showing
-)
bounded
+1 S3,
1 n
n e.
+]
n exists.a, actual
sequence lim

a, 2s
is
limn’ootherefore,
increasing limit
+
Real
Analysis is,
n+]
(n
+1)! the
That
n. that
all n,
monotonically
1+
>]+ n! convergent.,
all 3. show
fo
-I+ +
+... for and
<3
n. 3 2
=l+
| + all < be
between
<a,, can
for a therefore,
-| is
a, <a,> 2 It
since lies
, ,|> Rema
Also. is, Also, limit
a, Thus
114 that and the
so 3 that
by
 0. =a, inf lim =a,sup I:m
therefore and ),pOint
is cluster
2n =a,defined
by
e ne all for . <u,sequence
> the For 3.
=1. a, inf lim 3, =a,sup lim
h therefore and 3}. 2.1,points
{is cluster ofset the
n
n n=3,6,9,... 3+-,if
n
8,..=2,5, 2+-,ifn =a,
7.1+ifn=
. 1,4,
o setting defined
by <a> sequence the For 2.
-i. a,= inf im
s
and i. -and i are points chuter
= 4, p Iim therefore the For 1.
n,all for 1). defined
{- =a, by <a>sequence
cai the
IHustrations
limit)
of lower inferior
tor
by
a. inf lim denoted isand <a>
points
of cluster
the of I.saymember.
thcalled
e <a>
is ofset 10-2.
belinition
The
smcl sequence. bounded abe <a, Let
limit; superior
tor
a,Sup denoted isard <a,> of upper
lim by a
of U.member.
points
of cluster ofset the
greaenIiit thecalled <a>is 1-1.
hefinition
The
sequence. bounded abe <a, Let in
the rtudy ofeKt
section. present least
rcmber
snportaCe of
the Because
members,
these these of
f pnts
of cluster of e
2reates p1$e$$e
3 a,>
2s
weil rmermber oc east athas
Furtherrrsre.
we posntt. clustet
shon
tha have
al bunied bea ca,> Let
shownthat< alread, have scqucnce
e
SEQUENCE
INFERIOR LIMIT SUPERIOR
ANI) LIMIT 16.
OF
A
121 Sequene s