So, let us just recall we were looking at
set of real numbers, which is a complete ordered
field. So, on R there is addition, there is
multiplication and there is an order. And
the completeness meant there is R has Lub
or glb property. So, every non empty subset
of the real line which is bounded above has
got a least upper bound and similarly is every
non empty subset which is bounded below has
got greatest lower bound. So, that is called
the lub property and that is what makes it
a complete ordered field.
So, we were also looking at the notion of
Convergence of Sequences, I just wanted to
for the completeness sake wanted to say that
for every x, belonging to R, the order gives
rise to a notion of a distance. So, we define
what is called the absolute value of x to
be equal to x if x is bigger than or equal
to 0 minus x if x is less than or equal to
0. So, this is called this is called absolute
value 
of x.
And geometrically it signifies the distance
of the point x. So, this is 0 if you look
at the geometric representation x could be
here or x could be here. So, either way this
distance is mod x, okay. There are some obvious
properties which you would have come across
but anyway in case not you should revise those
properties. One, absolute value of x is always
bigger than or equal to 0, it is 0 if and
only if x is 0. So, that is positive definite.
Then second is alpha times x absolute value
is mod alpha times mod x, for every alpha
and x belong to real line. So, this is how
absolute value behaves with respect to multiplication,
okay. And the third how it behaves with respect
to addition. So, x plus y is not equal it
is at the most best possible is mod x plus
mod y it can be equal to sometimes, but not
always for every x and y belonging to R.
Fourth, this you might not have come across
but try to prove it mod x minus mod y is less
than or equal to mod of x minus y. So, here
is something if you have not come across check
it, why it is so, all these are useful properties.
So, let us go over to we were looking at Sequences.
So, we looked at a sequence a n, we said is
convergent if there exists some value call
it L, such that the distance between a n and
L or you can say that approximation, an is
an approximation to L can be made small, for
what? Such that this distance can be made
small for what epsilon, such that whatever
epsilon I give you I should be able to for
all n bigger than some n naught, and what
is n naught? There exists a stage n naught
such that this happens.
So, think it as a n is an approximation for
L and absolute value of a n minus L is the
error that you are making and that error you
should be able to make it as small as you
want, depending on how large and n is. So,
that should be for all n bigger than or equal
to n naught.
So, we looked at some examples, let us look
at one example was that the sequence minus
1 to over n is not convergent because it only
occupies the place 1 or minus 1, not only
it occupies only these places but occupies
very frequently every alternate times. Second,
we look at the sequence n is not convergent.
So, it does not come closer to any value but
keeps on become bigger and bigger.
Let us look at another one. So, this is equivalent
to saying if I look at the sequence 1 over
n, both are consequences and in fact equal
to what is called the Archimedean property
is convergent and the value is 0. Let us look
at one more probably before we go ahead. Let
us look at slight variation of the above,
minus 1 by n raised to power n, n bigger than
or equal to 1.
So, it is the same sequence 1 over n, but
now I am allowing it to fluctuate. So, n equal
to 1 the value is equal to minus 1, half minus
1 by 3 and so on, but still you can see that
it is coming closer and closer to 0. So, one
can try to prove it is convergent. So, try
to write a proof, try to write a proof of
this, okay. So, we will see what, what is
important, what is not important.
So, let us look at five. Okay, let us look
at the sequence n over n plus 1. Let us look
at this sequence, seeing all the above sequences
let us just observed a few things that this
sequence was not convergent, the first one,
because it was fluctuating, because it was
fluctuating at 1 and minus 1, this was not
convergent, because it was in some sense not
bounded. You cannot put some bounds that it
remains between some kind of barriers.
This is convergent to 0 we are going to look
at n is becoming larger so 1 over n is becoming
smaller. So,mething similar here, but in this
thing if you think of n is becoming larger,
but n plus 1 also is becoming larger, both
are becoming larger and larger. So, it is
not very straight forward to guess that whether
it is convergent or not convergent. But you
can think of that this there in some sense
n is increasing and n plus 1 they are increasing
at some same rate kind of a thing.
So, what do you expect? You expect that probably
it should converge, at the same rate, so,
they should cancel out, for enlarge the rate
should become same, of the ratio should become
same and that should be mean it should converge
to 1. So, there is a guess, that is how you
guess it is right. Now, try to prove it.
So, let us look at n over n plus 1. So, here
is where your thinking will come into picture.
If it is 1 over n, I can do something, numerator
is n, denominator is n plus 1, let us try
to make it same. So, I write this as n plus
1 minus 1 divided by. And why I am doing that?
The reason is this. So, that will make it
equal to 1 minus 1 over n plus 1. So, that
essentially looks like saying that is the
constant minus something and that something
is becoming smaller and smaller. So, you guess
that this converges to 1, so converges to
1.
So, let us write a proof of that. So, let
us look at a n minus 1, so what is that? That
is n over n plus 1 minus 1, a n minus L, I
want to make it small. So, what is that equal
to, so let us simplify n minus n plus 1 divided
by n plus 1 and that is nothing but, so, it
is 1 over n plus 1, is it okay? And that I
can make it small. So, give for every epsilon
bigger than 0 I can find n naught such that
1 over n naught plus 1 is less than epsilon.
So, that we will write so, we can write so,
this will imply that for every n bigger than
n naught, 1 over n plus 1, if n is bigger
than 1 over that is less than 1 over n naught
plus 1. So, that is n is bigger so n naught
plus 1, n is bigger. So, I am just trying
to write this thing for every n bigger than
n naught this also holds. Is that okay? So,
let us not spend time on that to write it.
You try to see something that this was convergent
right and that meant that the terms of the
sequence are 1, 1 by 2, 1 by 3 and 0, so everything
lies between 1 and 0. This was also convergent.
So, that also lies between minus 1 and 0,
okay, this is also convergent 1 so, it lies
everything is non negative, so, between 0
and 1. So, it seems to say if something is
convergent, that should be bounding. So, what
does that mean?
So, let us make it precise that is sequence
a n, so, we will say a sequence a n is said
to be bounded if there exist some alpha belonging
to R such that mod of a n is less than alpha
for every n, bigger than or equal to 1. All
the terms lie between some barriers, okay,
so that is bounded, so this is called bounded.
So, we should have a theorem based on our
experience of earlier results, that if a n
is convergent then it is also bounded, so
let us prove. So, what we want to do? We are
given it is convergent, so, what is given?
There exist number L belonging to R such that
for every epsilon bigger than 0, there exists
some n naught such that mod of a n minus L
is less than epsilon for every n bigger than
n naught, that is convergence.
So, let us try to draw a picture of this,
this is my L, this is L minus epsilon and
this is L plus epsilon. And what is this condition
saying, that a n lies here, a n naught lies
here, a n naught plus 1 that lies somewhere
here, so let us just probably, this is this,
this point, this point and so on.
So, for every n bigger than n naught everything
else is, so, what is outside this interval?
What terms of the sequence are outside, possibly
possibly we do not know some of them may be
inside, a1 maybe inside, it may be outside
a2 and so on up to a n naught minus 1 they
can be out they can be in.
So, maybe a1 is here, a2 is here, a3 is here
and so on. So, in any case, there are only
finitely many of them which are outside. So,
let us look at the smallest of a1, a2 up to
a n naught minus 1 and L minus epsilon. There
are only finitely many. So, let us write alpha
equal to minimum of a1, a2, a n naught minus
1, these possibly can be out. May be on the
left, may be on the right and but remaining
all are inside L minus, they are bigger than
L minus epsilon.
So, let us call this alpha and similarly look
at on the right side some of them can go beyond
L plus epsilon, only finitely many of them.
So, let us pick up the largest of them, so
let us call it beta equal to maximum of a1,
a 2, a n naught minus 1 and L plus epsilon.
So, then what should happen? So, then alpha
is less than or equal to a n is less than
or beta for every n bigger than n1. Is that
okay? So, everything else bigger than n naught
is inside the finitely many which can go out
or in we have taken the smallest and the largest.
So, in the picture if you look at it, so,
this will look like here is probably my alpha
and here is something like beta, alright,
and what is alpha? Alpha is the minimum and
beta is the maximum of this finite number
of things, whichever is the largest pick up
that.
So, that implies. So, this implies a n is,
is it clear that why is this implies a n is
bounded? See, our definition was something
slightly different. We said bounded when mod
of a n is less than, but here I am saying
a n is between alpha and beta, but that does
not matter. I can always have a number if
you like such that alpha and beta lie between
minus of that number and plus of that number.
So, mod of the a n will be between, is it
okay? Clear? Or you can pick up whichever
alpha is, mod alpha is bigger or mod beta
is bigger then all a n will be less, mod a
n will be less than or equal to that mod of
that number. So, both are equivalent ways
of saying.
Saying that a n bounded is this or all the
a n lies between some alpha and beta, the
two statements are equivalent ways of saying
a n is bounded, okay. So, what we have said
is, if see, now this is what is important
is if is convergent, then it is bounded. That
means boundedness is a necessary condition
for a sequence to be convergent, boundedness
is a necessary condition for a sequence to
be convergent.
So, if I given a sequence, the first thing
I should look at is whether it is bounded
or not. If it is not bounded, then it is not
going to be convergent, so problem is over.
If it is bounded, then I will go ahead. So,
all necessary conditions are useful in verifying
something is not true. If sequence is not
bounded, then it cannot be convergent, so,
that is the reason. In your calculus you might
have come across a statement probably will
prove it again later on, that you had done
calculus local maxima, local minima, of functions.
If a function is differentiable at a point
and that point is a local maxima, then derivative
must be equal to 0 that is also a necessary
condition. That for a differentiable function
if it is a point of local maxima or minima
the derivative is 0, that means, at that point
the tangent should be horizontal.
So, that is why when you want to analyze local
maxima or minima of functions, what you do?
You first look at derivative, if possible
derivative equal to 0. That is why that theorem
comes and those are possible points where
local maxima, minima can occur.
So, necessary conditions are useful things
even though they give something negative it
not true. So, same way, a n is convergent,
implies a n is bounded. So, not bounded will
mean it is not convergent. So, that could
be the reason for example, you can look at
that sequence, this is not convergent, because
it is not bounded. You can say that way or.
So, but this one is bounded but not convergent,
so, converse of the statement is not true.
A sequence may be bounded, but may not be
convergent and this is a obvious example of
that, okay.
So, let us keep that in mind and go ahead.
Now, how does one try to analyze whether a
sequence is convergent or not? Are there any
tools, one is obviously that if is not bounded
not convergent, so, that is not one of them.
Other possible tools are what are called algebra
of limits of sequences, we will not prove
all of them, we will just indicate. So, that
helps in analyzing limits.
So, so let us write theorem, let a n, b n
be sequences. So, okay, such that let us write
the conditions here also such that most sometimes
I will just write is equal this way b n converges
to m. So, this means the sequence a n is convergent
and the limit is L, sequence b n is convergent
and the limit is something m. Then the following
holds, one, see given a sequence a n, that
is a number, b n is a number. So, you can
add and nth terms of the two sequences.
So, you can form a new sequence, which is
a n plus b n, so that is a new sequence. So,
the theorem says if a n converges to L, b
n converges to m, then this must converge
to L plus m. So, this is algebra, we are adding
sequences. If we are adding sequences algebra
says you should be able to subtract also.
So, minus that also holds, okay. Can I multiply?
Yes, why not, a n, b n, nth terms we multiply.
So, let us follow our notation that this one
converges to L and we have written it as L.
So, a product of limits is, limit of the product
is equal to product of the limits, you can
interchange the two. Limit of a product of
sequences is nothing but the product of the
limits provided both converge. Difference
subtraction all are okay, you can okay. Let
us look at, what about division? I want to
say that a n divided by b n converges to L
by m. I want to say that, not possible?
Always, because obvious thing can go wrong
m could be 0. So, L by m does not make sense.
So, let us write condition, if m is not equal
to 0, at least then L by m makes sense, okay.
Now, what if m is not 0, what can you say
about b n? Because I am going to define a
n by b n, for that also all should also be
not 0. Or at least from some stage onwards,
they should not be 0, because I am looking
at the convergence, so I am looking at only
at the tail of the sequence. So, I should
be able to say that b n are not 0, if m is
not 0 from some stage onwards, can I say that?
Yes, Paul, we can do that.
Then let us write b n is not 0 for every n
bigger than n naught. And so I will indicate
why it is so because that is an important
thing. I will just indicate the proofs. And
that will help you to understand why and how
we are doing things. So, first one, I want
to look at a n plus b n. I know a n converges
to L, b n converges to m. So, this, sum minus
L plus m, I should be able to make it small.
But what do I know? I know something about
a n minus L, I know something about b n minus
m. So, I should separate them out, if possible,
somehow or the other.
So, this is what you want to reach somewhere,
keep the target in mind and somehow you have
to go in that direction, that is how problem
solving and writing a proof means. Not by
hook or crook by logical reasoning, okay.
So, this one, I can try to write at least
a n minus L plus b n minus m, and then these
two I know that I can say something about
them. And now here comes the triangle inequality.
It says this is less than or equal to plus
b n minus m. So, that makes everything work,
bn with the negative sign same thing will
work.
Let us look at product, what is happening
about the product? I am just trying to indicate
how the proof should go. I have to look at
a n, b n minus L m. Now, once again I know
only that mod of a n minus L can be small,
mod of the b n minus m can be small. So, how
do I separate out those things now, here a
n is multiplied by b n. So, I have to do some
kind of manipulation to separate out.
So, let us write this is equal to a n, b n,
I want a n minus L. So, let us add b n minus
L and add that also, add and subtract n minus
L m, so I have not change anything. Only I
have added and subtracted this because then
I can take out b n out and I will have something
a n minus L. So, my target is in mind, what
is it? So, this is b n, a n minus L plus L
into b n minus m. Now, at least I can separate
out and worry about it. So, mod of b n, mod
of a n minus L plus mod of L into mod of b
n minus m. So, at least I have separated out.
It is something like if you have a 200 rupee
note and you want to pay somebody 10 rupees
you have to get a change, you should have
that denomination with you otherwise you cannot
pay. Same way I want to use mod a n minus
L, mod b n minus L. Somehow I have to bring
in those things in my analysis.
So, now, I know that this will become smaller.
In intuitively this goes to 0 this goes to
0, multiply something with constant and that
is going to 0 so that will also will go to
0. Is it okay for everybody? Now, here is
a problem this b n may not go to 0 or b n
is not a constant. As n changes, I know b
n multiplied by something is 0, but this b
n is not 0. Yes, so here at times, that if
I can say it remains between a constant, it
does not blow up, it does not become infinite
kind of a thing.
So, boundedness, so it is less than or equal
to some bound, so let us call it is alpha
times a n minus L, where here I am using the
fact, the remaining thing. So, here one is
using that fact that if a sequence is convergent,
then it is bounded and everything is coming
out naturally that requirement is coming keeping
what I have in mind and what were I want to
reach. So, this will go. So, this will be
the second part proof essentially.
Let us look at the third part. So, we are
saying that b n converges to m which is not
equal to 0. So, let us draw a picture, here
is 0 and limit is either here or here, either
m is here or m is here on the left or on the,
we do not know where, positive or negative,
does not matter, but we know that b m is going
to converge. So, it is going to come closer
to the limit eventually, after some stage
onwards, the tail should be near the limit.
So, and I want to avoid.
So, let us take this neighborhood if it is
on right hand side. So, take a neighborhood
that means what? m minus epsilon and m plus.
So, choose epsilon such that m minus epsilon
is strictly bigger than 0, I can do that because
convergence says for every epsilon something
happens. So, a tail will come inside now,
a tail will come inside now here. That means,
all the b n after some stage will be bigger
than 0, if m is bigger than 0, is it clear?
So, that is a idea of limit, limit is not
evaluating at a point it says what is a behavior
of the tail of something. So, that is precisely
being used here, that if the limit is positive,
then the tail of the sequence will be positive,
for some tail and if negative then the some
tail similarly you can go on inside also.
So, a n by b n will be defined and then again
you have to manipulate a n by b n minus L
by m, LCM and do some things, arrange those
things, okay. So, do that.
By the way, as far as the examination is concerned,
this will not be asked, the proof, okay? But
try to understand what is happening, try to
write out a proof yourself that will help
you to understand mathematics behind this,
okay, do not worry about exams too much, try
to understand things.
So, what we are saying is, here are the, what
this is called the Algebra of Limits. Why
Algebra? Because I am adding, I am subtracting,
I am multiplying, I am dividing that is what
Algebra is. So, it says essentially that,
so this can, this helps now. How does it help?
It helps in analyzing limits of sequences,
because given is a n, I can break it into
probably smaller parts, which are either addition
of something or multiplication of something
or like that. For example, let us write some
example of this why it is useful.
