In today’s lecture, we are going to learn
some more tests, about convergence of an
infinite series, one is called the ratio test,
and other is called is root test; both are
wide
application in testing whether an infinite
series converges or not. And then we are going
to deal with series, which are not necessarily
of non negative terms. We are going to deal
with a particular kind of series called alternating
series, and try to understand the notion
of convergence for that kind of a series.
.
So, we first start with ratio test. The ratio
test is actually outcome of the comparison
test, we will see how it related to the comparison
test as follows. Let suppose I have a
n, which as the following property, that modulus
of a n plus 1 by a n, is lesser equal to
q. Where first of all, these quotient should
make sense, so I will say a n is not equal
to
0. And I put a condition 0 strictly less q
strictly less 1. And I assume, that this
inequality is true, eventually, that is after
one stage, this a n plus 1 by a n is lesser
equal
to q. So, in particular, what I mean is that
mod a n plus 1, divide by a n is lesser equal
to q.
.Let us say for all n, bigger than or equal
to some capital N, if this happen then,
summation a n, converges absolutely, this
is part a. Then the part b is, if modulus
a n
plus 1 by a n, is strictly bigger than 1,
for all n bigger than or equal to capital
N, then
summation a n diverges. So, let us see, once
again, what I say that suppose I have an
infinite series summation a n. I want to test,
whether it converges or not, suppose all the
a n is non zero, that means all the terms
are non zero.
Then, I look at modulus of a n plus 1 by a
n, if it happens than this quantities are
bounded by some positive quantity, which is
strictly less than 1 eventually than means
after some stage, then the series converges.
And, if it happens, that modulus of a n plus
1 by a n it is strictly bigger than 1 eventually
after some stage, then the series diverges.
Now, coming to the proof of it, it is very
easy to prove it you will see, it is just
simple
application of the comparison test. The easiest
part to prove b first, the given condition
says, that mod a n plus 1, is strictly bigger
than mod a n, if n is bigger than or equal
to
capital N.
In particular I can say, that mod a m, is
strictly bigger than mod a N for all m bigger
than or equal to capital N, this simply means,
that the coefficient a n of the infinite
series they do not converge to zero. Because,
if a n converges to zero, then after some
stage, they have to be less than epsilon,
for arbitrary choice of epsilon. In other
words,
they can be made arbitrarily small after some
stage. But here we see, it neither be, made
smaller than, modulus of a n, because it always
bigger than modulus of a n.
..
That implies the sequence a n does not converge
to zero, this implies, summation a n
does not converge. So, b is simple, now coming
to a, what is given to me, it is given
that modulus of a n plus 1 divided by a n,
this is lesser than or equal to q, for some
q,
which lies between 0 and 1. This q is a constant,
this does not depend on n, this is true,
for all n bigger than or equal to capital
N. Now, this then implies, that mod a n capital
N plus 1 is lesser equal to q times, modulus
a capital N.
And also, modulus a N plus 2, I can look at
that, that is lesser equal to, modulus of
a N
plus 1 into q, which is further lesser than
or equal to, q square into a mod, mod a N.
This way I will get finally, that mod a N
plus r is lesser equal to q to the power r,
times
mod a N, this is true, for all r strictly
bigger than 1.
..
What does this mean, that means the infinite
series, if I look at summation a n, then for
large n, mod a n is dominated by, a geometric
series of the form, summation mod a N
times, q to the power r. Because of my assumption,
that q lies between 0 1, 0 and 1, this
geometric series converges. Then by comparison
test, summation a n converges, it is
just a simple application of the comparison
test, but what fundamental here is, that q
lies between 0 and 1.
Now, once we have this, as a corollary, we
can have the standard statement of the ratio
test, which you can have many books on that,
it is enough form it is applicable. So, that
forms says, that summation a n I look at a
n is not equal to 0. And limit n going to
infinity, modulus of a n plus 1 divided by
a n, is equal to L. Suppose I have this, I
have
an infinite series a n, all the an’s are
non zero, and limit n going to infinity, modulus
of
a n plus 1 by a n exists, and it is equal
to L. Then if L is less than 1, this implies
summation a n converges, number 2 is, if L
is bigger than 1, then summation a n
diverges. But what happens if L is equal to
1, then the test is inconclusive, that is
the
series might converge, it may diverge also.
..
That is easy to see in the following example.
So, first we illustrate 3, so case 3 look
at
the series, summation 1 by n, also look at
the series summation 1 by n square, I know,
that this series does not converge 
and I know that this is converges. That is
a n here, a n
is equal to 1 by n. And here, a n is equal
to 1 by n square, then what is a n plus 1
by a n,
that is, n by n plus 1, here a n plus 1, by
a n that is equal to n square. So, what is
then
limit, n going to infinity a n plus 1 divided
by a n. This is same as, limit n going to
infinity n by n plus 1, that is limit n going
to infinity 1 by n plus 1 by n, which is equal
to 1.
And here, so you see, in both the cases, the
required limit of a n plus 1 by a n, turns
out
to be 1. But in the case, the series converges,
and in the 1 case the series does not
converge. That means, the case L is equal
to 1 . does not reveal
any about the infinite series. So this test,
will only in the cases, when L is less than
1, or
L is bigger than 1. L is less than 1 implies,
the series converges that is, what is given
in
1, and L is bigger than 1, the series a n
diverges.
..
Now let us see a quick proof of this, so we
start with 1, given that limit n going to
infinity, mod a n plus 1 by a n is equal to
L, which is less than 1, I know that. Now
choose, epsilon bigger than 0, such that L
plus epsilon is less than 1 and L minus
epsilon bigger than 0. Then by the definition
of convergence of sequence, there exist
capital N. Such that for all n bigger than
or equal to capital N this quantity is less
than
L plus epsilon and bigger than L minus epsilon.
Now, I just concentrate on this part,
this is true for all n bigger than or equal
to capital N.
.
.So, according to our previous result, which
I have proved, this quantity l plus epsilon,
can be considered as q. And, since my choice
L plus epsilon is less than 1, this q is less
than 1 this implies summation a n converges
by the previous result. The previous result
was, if after some stage, modulus of a n plus
1 by a n is less than q, where q is less than
1 then the series converges. Here, I got the
exactly the same thing, instead of q i got
epsilon, but which is less than 1, this can
be considered as q, and hence the previous
result applies.
Now, let us come to the second part, this
says, the limit n going to infinity, modulus
of
a n plus 1 by a n is equal to L, which is
strictly bigger than 1. What I do is, choose
epsilon, bigger than 0, such that capital
L minus epsilon, strictly bigger than 1. Since
L
is strictly bigger than 1, I can always choose
an epsilon, such that L minus epsilon
strictly bigger than 1, and epsilon is positive.
.
Then by the definition of limit, this implies
modulus a n plus 1 divided by a n is bigger
than L minus epsilon, for all n bigger than
or equal to capital N. But, then I can
compare this, again with the previous result,
that as l minus epsilon is strictly bigger
than 1, the series diverges, summation a n
does not converge. Look at the previous
result, where I have proved ., that this is
the statement, I am
looking for…
..
See if n is bigger than or equal to N and
modulus of a n plus 1 a n is bigger than 1
for
all n bigger than or equal to N, then summation
an diverges. It does not converge,
because the term do not go to zero, that is
what is happening.
.
Now, the situation I have at hand is precisely
this, that modulus of a n plus 1 a n, this
is
bigger than L minus epsilon. But, L minus
epsilon by my choice, strictly bigger than
1,
then the series, summation a n does not converge,
because it is terms do not go to zero,
that gives you the ratio test.
..
Let us look back again at the statement of
the ratio test, it says what that first, have
six
series summation a n, the k has to be taken,
all the terms are non zero. Otherwise, it
is
difficult to apply the ratio test, because
we are dividing by something. So, if all the
terms are non zero, I look at modulus of a
n plus 1 by a n, and then I look at the limit.
Suppose the limit exists, then I have to check,
whether limit is less than 1, if it is then
the series converges. If the limit is bigger
than 1, then the series diverges, if L is
equal
to 1, then we cannot say anything, and we
have to try for something else.
Now, the next test, what we are going to deal,
with is called the root test ((Refer Time:
18:05)), this is also consequence of the comparison
test, but here the comparison being
made with geometric series. For example, suppose
the situation is this, that I have
summation a n I know that all the an’s are
bigger than or equal to 0. And let us say,
0
lesser equal to a n is less than or equal
to x to the power n, where 0 less x less 1.
Then,
it is very clear from the comparison test,
that the series summation a n converges,
because the geometric series summation x to
the power n converges.
This implies, if my condition was, that 0
lesser equal to a n to the power 1, by n lesser
equal to x, where 0 less x less 1, then summation
a n converges. Because, a n to the
power 1 by n lesser equal to x, it would imply
an is lesser equal to x to the power n,
then I can compare the root test is essentially
the same thing, written in different
language in terms of limit.
..
So, let me first, give you the statement.
So, I write it as a theorem, summation a n
is
given, and limit n going to infinity, modulus
of a n whole to the power 1 by n is equal
to L. Then, if L is less than 1, then the
series converges, if L is bigger than 1, then
series does not converge. And third is, like
the ratio test, if L is equal to 1, the test
fails,
that is no conclusion, may be drawn. Again
as an illustration of 3, what we have to do
is take an to be equal to 1 by n. Then limit
n going to infinity, which is certainly, is
equal to 1.
But I start with a n is equal to 1 by n square,
then also, this limit is also is equal to
1,
but summation 1 by n diverges and summation
1 by n square converges. So, in this
case, L is equal to 1, and hence the test
is not conclusive. So again, we just need
to
prove the case 1 and case 2.
..
So, let us come to the proof, what is given
to me it is given, I limit n going to infinity,
modulus of a n, whole to the power 1 by n
is equal to L and L is less than 1. Choose
epsilon again, such that L minus epsilon L
plus epsilon is strictly less than 1, this
then
would imply, that there exist, some capital
N. Such that, for all n bigger than or equal
to
N, modulus of a n to the power 1 by n, is
less than L plus epsilon ..
It imply, that modulus of a n is less than
L plus epsilon whole to the power n, for all
n
bigger than or equal to capital N.
Now, notice that L plus epsilon is strictly
less than 1, so L plus epsilon to the power
n, if
I look at the sum, that gives me a geometric
series, which converges, and modulus of a
n
is less than that, by comparison test. Then,
summation a n converges. In fact, a n
converges absolutely, so in this case, we
again comparing, and again with geometric
series.
Now, the second case, the limit n going to
infinity, modulus of a n to the power 1 by
n,
that is L and L is bigger than 1. Now choose,
epsilon bigger than 0, such that L minus
epsilon is still bigger than 1, since L is
bigger than 1, I choose some such epsilon.
..
This implies then, by the definition of limit,
that there exist capital N such that, for
all n
bigger than or equal to N. Modulus of a n
to the power I by n is bigger than L minus
epsilon, this implies modulus of a n is bigger
than L minus epsilon, whole to the power
n, for all n bigger than or equal to capital
N. Now, notice L minus epsilon being bigger
than 1, L minus epsilon to the power is always
bigger than 1, this would then imply, that
modulus of a n is bigger than 1, for all n
bigger than or equal to capital N. This certainly
implies, that a n does not converge to zero,
this implies, that the summation an does not
converge.
.So again the statement is very simple, what
I do is, a n is given, I look at modulus of
a
n to the power 1 by n, and I just calculate
that limit, If that limit, is less than 1,
then the
infinite series converges, if that limit is
bigger than 1, then the infinite series does
not
converge. But, if the limit is equal to 1,
then I have to careful, because in that case,
infinite series may converge, and it may not
converge.
.
Now, let us see, some examples here, let us
look at summation, so this is the first
example, 1 by log n whole to the power n,
n from 2 to infinity. So, here a n is equal
to 1
by log n whole to the power n, that means,
a n to the power 1 by n, that is 1 by log
n, this
goes to 0, as n goes to infinity. So, in this
case, L is equal to 0, which is less than
1, that
implies the series converges, similarly, I
can look at summation n by n plus 1 whole
to
the power n square. So, in this case, a n
is equal to 
that means, a n to the power 1 by n
trans out to be, which is same as…
..
So I can write this as, and since we known
that limit n going to infinity, 1 plus 1 by
n
whole to the power n, is e which is strictly
bigger than 1, this then implies, since I
have 1
by it implies, that limit n going to infinity,
a n to the power 1 by n, this is 1 by e, which
is
strictly less than 1, this implies then the
series converges. So, given an infinite series,
now we have certain techniques, which we can
apply to test whether the series converges
or not.
The first one, the most effective 1, is the
comparison test, you try to compare it with
geometric series, or summation 1 by n to the
power p. Then we have the limit
comparison test, then we have the ratio test,
now we have seen root test, after this, I
am
going to look at series, which are not always
series of non-negative terms, that means,
here, negative terms can also come. And, we
want to look at certain test, which will tell
me the convergent of this kind of series.
...
Now, we come to something called alternating
series, so what is an alternating series,
suppose I have a sequence a n, and a n is
bigger than or equal to 0. Then, I look at
the
infinite series of this form, summation n
from 1 to infinity, minus 1 to the power n
plus
1, times a n. Notice then, what is the first
term of the infinite series that means, when
n is
equal to 1, this is a 1. So, the series looks
like, the first term is a 1, then I take n
to equal
to 2, that means, minus 1 to the power 3 times
a 2 minus 1 to the power 3 is minus 1, so
it is minus a 2.
Then comes, plus a 3 minus a 4, plus a 5 and
so on, So, you see, the signs actually
alternate 1 to another, first term comes with
the positive sign, second term comes with
the negative sign, third comes with the positive
sign, fourth term comes with the negative
sign and so on. That is, either terms alternating
series, all we are interested, in that under
which criteria, this kind of a series converges,
the particular example we have in mind is,
this series, summation n from 1 to infinity,
minus 1 to the power n plus 1 times 1 by n.
So, the series look likes, if I look at first
n is equal to 1, the first term is 1, then
minus
half, plus 1 third, minus 1 fourth, plus 1
fifth and so on. Notice that, this series
is not
absolutely convergent, because if I take modulus
of the terms, all I get is summation 1 by
n, which I know does not converge. But, the
question still remains “does the series
converge”, because absolute convergent implies
convergent, but not the other way, so we
.are interested to know, that this particular
series converges or not, that we will get,
by
something called Leibnitz test.
The statement is like this, suppose a n is
non-negative for all n, and the sequence a
n is
decreasing, and a n converges to 0. So, what
are the properties, I have a sequence a n,
such that elements are non-negative, it is
decreasing, and it converges to 0.
.
.
Then, the alternating series summation n from
1 to infinity, minus 1 to the power n plus
1
a n converges, before we go to the proof of
this, come back to the example, 1 by n is
bigger than or equal to 0, 1 by n plus 1 is
less than 1 by n, this is the condition
decreasing, and also 1 by n goes to 0. Hence,
by Leibnitz test, it would imply, that
summation n from 1 to infinity, minus 1 to
the power n plus 1, into 1 by n, converges,
which is not absolutely convergent.
So, in particular it is also gives me an example
of a series, which converges, but not
absolutely convergent. So, the properties
are very simple to remember, for this Leibnitz
test as a model, you should always remember
the series, minus 1 to the power n plus 1
times 1 by n. The conditions are exactly analogues
to this series, that means, a n are nonnegative,
decreasing and goes to 0, as n goes to infinity,
then the alternating series
converges.
.Now let us come to the proof of this, so
first we will look at, partial sums of the
series,
but the even partial sums, and the odd partial
sums. So, we want to consider, the odd
partial sums s 2 n plus 1, and s 2 n, so what
are the definitions, very simple, usually
s m
means, summation n from 1 to m minus 1 to
the power n plus 1 into a n.
.
Now, let us see, what kind of sequences this,
this partial sums are, let us first start
looking at, s 2 n, whether it is increasing
or decreasing or what. So, first let us check,
s 2
n plus 1, minus s 2 n, what is this, this
is summation i, from 1 to 2 n plus 2 minus
1 to the
power i plus 1 a i, minus summation i from
1 to 2 n, minus 1 to the power i plus 1 a
i.
Once, I write this, many terms cancel except,
the last 2 terms that means, what remains
is
minus 1 to the power 2 n plus, that means,
the terms i is equal to 2 n plus 1.
So, it is 2 n plus 1 plus 1, that is 2 n plus
2 a, 2 n plus 1, then the next terms plus
minus 1
to the power 2 n plus 3 a, 2 n plus 2, that
means, what I get is, a 2 n plus 1, minus
a 2 n
plus 2. Now, I know, that the sequence a n
decreasing, that property given to me, that
means, a 2 n plus 2 is less than a 2 n plus
1, that means, this is bigger than or equal
to 0.
What does this prove, this implies, the sequence
is 2 n is and increasing sequence,
because I have seen, then the next term is
bigger than the previous term, and that is
happening for each n, so the sequence is increasing.
Now, if I start with, the odd sub-sequence
of partial sums, the odd sequence of partial
sums is 2 n plus 1. Then I would certainly
look at s 2 n plus 3 minus s 2 n plus 1, then
the
.calculation exactly like the previous one
tell me, that what remain is, minus 1 to the
power 2 n plus 3 a, 2 n plus 2 plus, minus
1 to the power 2 n plus 4, a 2 n plus 3, that
means, a 2 n plus 3, because minus 1 to the
power 2 n plus 4 is 1, minus a 2 n plus 2,
but
notice, that the sequence is decreasing, that
means, a 2 n plus 3, is less than a 2 n plus
2,
which is less than or equal to 0.
So, the behavior changes, this implies then,
the sequence s, 2 n plus 1, is a decreasing
sequence. So, the even sub-sequence is increasing,
that is what i got, and the odd subsequence
is decreasing, fine. Now, I want to compare
between s 2 n and s 2 n plus 1,
what is the relation among them.
.
So, let us look at, s 2 n plus 1 minus s 2
n, if you write down this quantities, you
will see,
what will get is, minus 1 to the power 2 n
plus 2, a 2 n plus 1, that is a 2 n plus 1,
because
2 n plus 2 is an even number. So, the minus
1 to the power 2 n plus 2 is any way 1, this
is
bigger than 0, because all the terms are non-negative,
so this implies, s 2 n plus 1, is
bigger than or equal to s 2 n. Now, I have
two more information, about this s 2 n and
s 2
n plus 1, I have already seen.
So, let me write here, s 2 n increasing, and
s 2 n plus 1 is decreasing, now notice one
thing, i can write it in this form, s 2 n
plus 1, that is bigger than or equal to s
2 n, I got
that. Since, it is increasing, it is certainly
bigger than or equal to s 2, because s 2 n
is
increasing, so all the higher terms are bigger
than the first term, the first term is s 2.
So, s
.2 n is bigger than s 2, on the other hand,
if I look at s 2 n plus 1, that is decreasing,
that is
all the terms less than or equal to the first
term. So, that means, which is less than or
equal to first term, which is s 1, and this
is true, for all n, this implies, then that
the s 2 n,
is bounded increasing sequence, similarly
s 2 n plus 1, is a bounded decreasing sequence.
.
But, we know, that every bounded increasing
sequence, and bounded decreasing
sequences, they converge to some numbers,
using that this implies, that s 2 n, converging
to some number s, and s 2 n plus 1, that converge
to some number s prime let us say.
Because, these are bounded increasing and
decreasing sequences, now this implies, now
that if I look at s prime minus s, that is
limit n s 2 n plus 1, minus limit n s 2 n,
which is
same as, limit n s 2 n plus 1, minus s 2 n,
that is limit n, a 2 n plus 1.
Now, there is another condition on that an’s,
that as n goes to infinity, a n goes to 0,
as n
goes to infinity a 2 n plus 1 is also goes
to 0. Because it is a sub-sequence of the,
sequence of an’s, so this is 0, this would
then imply, that s is equal to s prime, but
since,
s is equal to s prime, this would imply, that
the sequence of partial sums s n if I look
at,
converges to s, which is same as equal to
s prime, this simple fact about the sequences,
which we have used earlier also that if we
have the sequence s n, look at it is even
sub
sequence, and look at its odd sub sequence.
If, the even sub sequence, and odd sub sequence
has the same limit, then the whole
sequence, also has the same limit, in particular
it converges, by that observation. Now,
.sequence of partial sums of the alternating
sequence s n converges to a number s, this
implies the whole series converges.
.
Now, i am going to tell you about, another
test, this time without proof, but it comes
very
handling and many practical situations, you
have to use that test, it is called the Dirichlet
test. Suppose, a n and b n are 2 sequences,
if number 1, that summation n from 1 to m
a
n, which i call s m is bounded, that means,
the partial sums of a n is a bounded set.
Number 2 is b n is decreasing and b n converges
to 0, then summation over n, a n b n
converges, this is called Dirichlet test.
Let us, elaborated on condition one, what
we mean is, we look at s 1, which is just
a 1,
then we look at s 2, which is a 1 plus a 2,
then you look at s 3, that is a 1 plus a 2
plus a 3
and so on, so that s n is equal to a 1, plus
a 2, plus a 3 up to a n look at this numbers.
Then, partial sums are bounded, which means
there exist, some m bigger than 0, such
that, modulus of s n is less than or equal
to m, for all n. This is what we mean, by
saying
that s m is bounded, that means there exist
a number capital M, such that, in the modulus
the partial sums of a n are less than or equal
to, that number M. If that happens, bn’s
are
decreasing, and b n is going to 0, then the
series summation over n, a n, b n converges
this is called Dirichlet test.
..
So, let us just, illustrate it by the following
example, look at the following series,
summation n from 1 to infinity, cosine of
n theta divided by n, where theta is fixed
number, and theta is not equal to 0. Notice
that, if theta is equal to 0, then this series
does
not converge, because cosine of n theta is
1, and summation 1 by n does not converge,
but if theta is non-zero, then we are going
to prove, this series converges so what we
do
is, we want to apply Dirichlet test here.
So, I put a n, is equal to cosine n theta,
and b n is equal to 1 by n, now bn’s are
decreasing and decreasing to 0, that is clear.
So, this is decreasing, and b n goes to 0,
now all I need to prove is that summation
n from 1 to m, cosine n theta, if I look at
the
mod, it is less than or equal to some number
capital M for all m. So, essentially now,
i
am bother about this kind of a finite sum,
that is cosine theta, plus cosine 2 theta,
plus up
to cosine n theta. I want to show this is
lesser equal to capital M.
What I do is I view cosine theta as, the real
part of the complex number, and that is very
well complex number all of us know it, if
I look at e to the power i theta. This is
cosine
theta plus I sin theta, and we also know by
de moivre formula, that e to the power i n
theta, is cosine n theta, plus i sin n theta.
Now, it is always it is true, that real part
of the
complex number in the modulus is always less
than or equal to modulus of the complex
number.
.So, this, then implies, the modulus of cos
theta, plus cos 2 theta, plus up to cosine
n
theta, is certainly less than or equal to,
modulus of e to the power i theta, plus e
to the
power twice i theta, plus up to e to the power
i m theta. Now, notice the right hand side
sum, I can actually calculate, because it
is a geometric series, with common ratio e
to the
power i theta.
.
.
So, I am going to use that, e to the power
i theta, plus e to the power twice i theta,
plus
up to e to the power i m theta, that I can
write as, e to the power i theta, into 1,
plus e to
the power i theta, plus e to the power i m
minus 1 theta, which then by a well known
formula, g p series is 1 minus. Now, this
implies then, that modulus of e to the power
i
theta, plus up to e to the power i m theta,
is modulus of e to the power i theta, into
1
minus, by the previous formula.
Now, since modulus of e to the power i theta,
or e to the power i n theta is always 1,
what I get is, this is lesser equal to using
the triangle inequality 2 divided by modulus
1
minus e to the power i theta. The point is,
now think are independent of m, this is true,
for all n, this then imply modulus of cosine
theta plus cosine 2 theta plus cosine m theta
is lesser equal to 
this well defined as theta is not equal to
0.
And hence the Dirichlet test applies, . I
go back to the previous step,
this m I have actually found out, this m Trans
out to be 2 divided by modulus of 1 minus
e to the power of i theta. It depends on theta,
it seems, that is it should not depend on
.little m’s, which it does not, so the partial
sums of an’s are bounded, bn’s decreasing
and
going to 0 this then implies by Dirichlet
test, that the series converges. This is all
we had
to cover in the infinite series. In the next
lecture we are going to talk about something
called power series.
.
