Welcome to a video on the Limit Comparison Test.
for the convergence of an infinite series.
We need to be careful not to confuse this with the
Comparison or Direct Comparison Test.
This is the Limit Comparison Test.
So, limit comparison test states that we have two infinite series
that both have positive terms,
and
the limit of the quotient,
of a sub n and b sub n
equals L,
or L is finite and positive,
then both series converge or diverge.
The idea behind this test is,
if we let a sub n be the given series,
and that resembles b sub n,
and we know b sub n, either converges or diverges.
then the limit of the quotient equals
L which is finite and positive,
we can use that,
to determine whether,
a sub n also converges or diverges.
So, the limit comparison test can be helpful when,
maybe we can't apply the direct comparison test.
Let's go take a look at some examples.
This series here should remind us,
of the infinite series of three divided by four to the power of n,
Notice as, n increases we multiply the denominator by additional factors of four,
which is the same as multiplying by one-fourth each time.
So, this is a geometric series, where R is equal to one-fourth,
and therefore converges by the geometric series test
So, R is equal to one-fourth, which is less than one,
and therefore converges.
Now, we can use the limit comparison test,
to find the limit of this quotient,
and see if L is positive and finite,
and if it is,
we know that the given series is also convergent.
so now we'll take the limit,
as n approaches is infinity,
of a sub n divided by b sub n,
Now, doesn't matter which are these formula we let
equal a sub n and b sub n.
Let's go ahead that, this equal a sub n,
so three divided by,
four to the power n all divided by
b sub n which led equal the
given formula three
divided by four to the power of n minus three,
Let's go ahead and convert this to multiplication problem.
So, we'll have three,
divided by four to the power of n,
times a reciprocal of this. So, we'll have four,
power of n minus three.
divided by,
three.
Well here, the three simplify out,
We'er left with the limit as n approaches infinity,
four
to the n minus three divided by four to the n
and since we are dividing by one term, we can break this up to two different fractions
So, four to the n divided by four to the n,
minus three,
divided by four to the n.
Well, this is gonna equal one,
and as n increases our denominated increases without bound.
So, this will approach zero,
so one minus zero would equal one.
Since, the value this limit is positive,
and finite,
by the limit comparison tests,
the given serious,
also converges.
Let's go and take a look at another example.
Here we have the summation of five divided by the square root of,
n squared minus one,
So, if we ignore this minus one, we'd have the squared root of n squared
which is equal to n.
So, the series
resembles the harmonic series,
one divided by n,
which we know diverges by the p-Series test.
with P equal to One,
which is less than or equal to one.
So, since the series, it resembles diverges,
if we take the limit of this quotient,
where L is a finite positive number
it will also show that the given series,
diverges.
let's go to give it a try,
we have the limit,
as n approaches infinity,
of the quotient of a sub n,
divided by b sub n,
and again it doesn't matter which of these formulas,
let equal a-sub n or b-sub n.
For this problem, we go ahead and let a sub n equal five
divided by the squared root n squared minus one
divided by
b sub n, which is one divided by n,
Now, rewrite this as a product of two
fractions so we'll have the limit,
as n approaches to infinity of five,
divided by,
square of n square minus one, times
the reciprocal one over n, which is n
divided by one.
nothing simplifies here, so go ahead just multiply across the top and bottom
we'll have five n in the numerator,
divided by the square root a
n square minus one,
Remember! the technique to help us evaluate this limit would be to
divide by the highest power of n in the denominator,
but we do have an n squared in the denominator,
but its underneath the square root
and the square root of n squared is equal to n
So we gonna divide the numerator by n, and by the denominator by squared root of n squared.
So, our numerator will be five n divided by n,
into the denominator we're gonna have,
n squared,
divided by n squared,
both these underneath the square root,
minus one divided by n squared
this will give us a five in the numerator
and we'll have the square root of one,
minus one over n squared,
So, as n approaches infinity, we'll have five over the square one which is equal to five.
Well, this limit is positive in finite
so by the limit comparison test,
the given series also diverges.
I think, we have time for one more.
Take a look at the formula for the series,
there's a lot going on here but if we consider just the highest power of n
in the numerator denominator we have n to the fourth,
and then n to the six, in the denominator,
So, if we take n to the fourth,
over n to the six, and simplify,
we're gonna have, one over n squared.
So, we gonna say that this series resembles,
the series
generated by,
the summation of one divided by n squared
and this series converges
by the p-series test,
with p equal two,
which is greater than one.
So, now if we take the limit of this quotient, and its positive and finite
since this series converges, we can show that the given series converges.
Let's go ahead and give it a try,
will have the limit,
as n approaches infinity,
a sub n divided by b sub n
let's go ahead that this formula equal a sub n,
so have n to the fourth,
Plus two n squared plus one,
divided by,
six n to the sixth plus four n to the fourth.
Now, instead of dividing by,
one over n squared.
Let's go ahead and multiply by the reciprocal which should
be, n square over one.
Now, we'll multiply the numerator by n squared,
So, we'll have n to the sixth,
plus two n to the fourth,
plus n squared.
allover,
six n to the sixth,
plus,
four n to the fourth.
So, dividing each term by n to the sixth now,
Remember, for limits at infinity,
if the,
degree the numerator denominator are the same,
then the limit will equal the ratio of the leading coefficient.
So. this limit will be equal to one-sixth,
which,again is positive and finite,
so, since this series converges,
so does the given serious,
by the limit comparison test.
That'll do it for the limit comparison test.
I hope you found this video helpful.