 
Hello students, I Dr. Gajendra Purohit
Today we are going to discuss sequence and series
in that we will talk about
what is a comparison test, and with that how
we come to know series is convergent or divergent
Previously we discussed about the basic concept of convergent
and divergent
we discussed the basic terminology
and about the formation
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So students, look here what is the comparison test
If we have any sequence
or series
 
here the positive term is very important
 
 
 
 
and k should be finite positive
another point is
 
 
ultimately both the points are almost the same only
 
if we want to calculate the convergence of un
we will decide a vn
we will take vn depending on un
vn is a series
on vn we will test, on that ratio
if its value is finite
then we will talk about what is the behavior of vn
if it is convergent then our series
will also be convergent
if this value is finite and if our vn is divergent
then our un is also divergent
then you will have a question, that this vn
how we will know it is convergent or divergent
so, student, there is a P-series test
or helping test
here if the vn has this type of series
 
 
here, the p is power
and if it is greater than 1
then the series is convergent
if the power is less than 1 then it is divergent
 
what does this mean?
if you are given any question in an exam
there will be a sequence, this is un
 
looking at the un you have to take the n term of vn series
 
then you will check its ratio
that the value is finite
then with this helping test, we will come to know
if vn is convergent then un is also convergent
and if it is divergent then un will also be divergent
but for that, this value should be finite
let us take an example to understand
this concept and how to implement it
so as you can see
the value of un we have as
 
now we will see how we will calculate vn
 
 
let us take one more example
 
this is given
we will see the highest power of n
we will calculate its difference
then our vn will be 1/n
if in case we have
 
so here depending on un
you will choose vn
so here it is coming
 
 
now what we will do
we will check its ratio
 
 
 
 
 
 
put the values as follows
we will simplify it
 
we will check its limit
 
taking n common
solve as follows
 
 
we will keep n as infinite
value is as follows and it is positive
that is it is finite
and positive
it means we are implementing the test here
now we will see
this vn is convergent or divergent
we need to know the behaviour of vn
 
you can take it as shown
now we will see
 
we will check vn here
 
 
we have power p
here it is 1
 
by the helping test we conclude
 
 
 
 
 
the vn is
divergent
and if are vn is divergent then
by comparison test
un is
divergent
so students let me explain you one more time
first of all you will be given a sequence
take the n term of that
 
 
depending on this we will
calculate its vn
we will calculate limit
 
and if the value is finite then we will check
by helping test that vn is convergent or divergent
and here it is coming divergent
series is as follows and value of p is 1
it is coming 1
and at 1 it is divergent
so our vn will be divergent
and if vn is divergent then un is also divergent
if this value is finite
then we will decide on vn
if it is divergent then un is also divergent
 
let us take one or more examples here
so students check out one more question here
 
we have to test the convergent
so first we will check it's un
 
 
solve as follows
 
 
 
we will check what value we are getting
 
 
you will have to check
how to take its n term
you just have to solve it like this
 
these are odd
so odd is 2n+1 or 2n-1
 
 
in place of n we will keep 1
now what we will do
we have discussed how we will choose vn
there 3n in denominator
and numerator it is 1
so it will be
 
solve as follows
this needs to be kept in mind
now what we will do
we will calculate limit
 
 
so the value of un is
 
 
 
 
 
we will divide it
 
we will take it common
 
 
 
 
 
 
we will put the limits then the answer will be
it will be 2, which is finitie and
positve
as it is finite and positve which means
now we can see it is convergent or divergent
what we will do is
 
we will check by using helping test
 
 
when p is greater than 1 then the series is convergent
 
 
 
 
 
 
 
 
 
so students, this given series we have is convergent
so this is the way we will check it
let us see more questions on the topic
so students look at these two questions
you have to check that
this series is convergent or divergent
so what we will do is
 
 
 
in these type of questions, we will first rationalize
in order to remove its under root
 
 
 
 
so after rationalizing
 
simplify it as follows
 
 
 
we will be left with
we will choose vn
we have discussed the method
you have to check the degree of n
we will solve as follows
so its value will be
now we will check
the limit
 
 
 
 
 
solve as follows
 
 
 
 
 
 
we will simplify it
 
 
 
where there is n we will keep it infinite
 
 
 
now we will check vn is convergent or divergent
 
 
 
 
 
 
 
 
so un will also be convergent
 
 
let us take one last question
we are given the n term
 
you have to find out the given series
is convergent or divergent
whose n term is given
so now what we will do
we have a formula sin series
what is it let us see
 
 
 
 
simplify as follows
 
if we talk about this term
so here the degree of n is
 
we have to see the lowest degree
 
 
now we will calculate limit
 
 
we will divide this
 
 
 
 
 
 
 
solve it as follows
 
 
 
 
simplifying it
 
 
 
 
 
 
after putting limit this will be 0
then our value will be
which is finite and positive
now we will check
 
 
 
 
 
 
series vn is convergent
so by comparison test un is convergent
 
so, students, I hope you understood this topic
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