Hi, everyone.
Welcome back to integralcalc.com.
Today we’re going to be talking about how to use the limit comparison test to determine
whether or not a series converges or diverges.
And in this particular case, we’ve been asked to find the sum of the series from 2
to infinity of 10n squared divided by the quantity n cubed minus 1.
So with the limit comparison test and I’ve taken the screen shot here of the rules that
you’re going to use with the limit comparison test, this is from the table of convergence
tests that I have on my website.
But what we need is two series, one is our original series, a sub n and we’ve already
got that here.
But we need a second series called b sub n that we’ll take from pieces of a sub n and
we need to know that both of those series are greater than zero for all values of n
and that the limit as it approaches infinity of our original series a sub n divided by
the modified series b sub n is going to be greater than zero.
If those things are true, then we can conclude that whatever happens to b sub n also happens
to a sub n.
So if all those things were true and b sub n converges, then a sub n also converges and
similarly if all those things were true and b sub n diverges then a sub n is also going
to diverge.
So the first thing we need to do is figure out what b sub n is going to be.
What we want to do is we want to take the terms of a sub n that are going to have the
greatest effect on the magnitude of the function.
In other words, the terms that are most meaningful.
For example, if we plugged in the number 10000 for n to our original series a sub n, then
you can see here that 10n squared is going to have a big effect, n cubed is going to
have a big effect.
But you’ll have these huge numbers and then negative one here is going to be a tiny, tiny,
tiny insignificant piece of this fraction compared to the other two terms.
So, that means that 10n squared and n cubed are going to have the largest effect on the
magnitude.
We don’t have to consider the coefficient here, 10.
We just want the variables.
So we’re going to say that b sub n is going to be equal to n squared divided by n cubed
which we can simplify to be 1/n.
And that is actually the harmonic series which we know, diverges.
That’s just something that is a given that we remember from sequences and series.
So we’ve got a harmonic series and that will be b sub n.
So now, we need to make sure that or all n, a sub n and b sub n are always going to be
greater than zero.
Well, we know that n begins at 2.
So we’re interested in all numbers 2 and greater.
You can imagine here for a sub n, if we plug in the number 2, we’ll get a sub n of 2,
we’ll get 10 times 2 squared which is going to give us 40 in the numerator.
And then in the denominator, we’ll get 2 cubed which is 8 minus 1 is 7.
We get 40/7.
You can imagine that if we continue plugging in 3, 4, 5, 6, positive numbers, the numerator
is always going to be positive and the denominator is always going to be positive.
So a sub n is always going to be greater than zero.
Similarly, for b sub n, if we start at n equals 2, we get 1/2.
And then if we plug in 3, we get 1/3.
We plug in 4, we get 1/4, etc.
That’s always going to be a positive number.
Nothing we could plug in that’s greater than 2 is going to make this a negative value.
So we also know that b sub n is also going to be greater than zero.
So we can see that both of those things are true so we can check off this that a sub n
is greater than zero and b sub n is greater than zero.
Now, we need to find L.
We’re going to take the limit as n approaches infinity of our original series a sub n divided
by our modified series b sub n and whatever we get as a result is going to be L. So we’re
going to take the limit as n approaches infinity.
Now that we’ve shown that both a sub n and b sub n are greater than zero, we’re going
to proceed with the calculation.
So the limit as n approaches infinity of a sub n which is our original series so we’ve
got 10n squared divided by n cubed minus 1 and we’re going to divide that whole thing
by b sub n which we know to be 1/n.
So now, we want to just go ahead and simplify this before we try plugging in infinity.
So instead of dividing by this fraction here, we can multiply by its inverse.
Hopefully you will remember that from algebra that when you divide by a fraction, you can
instead multiply by the inverse.
So instead of dividing by 1/n, we’ll multiply it by n/1.
So when I combine those two fractions, I’ll get 10n cubed over n cubed minus 1.
So now from this point, I can go ahead and divide thru the entire fraction by n cubed.
Remember that this is a useful tool when you’re trying to evaluate an infinity when you have
a limit as n approaches infinity or whatever your variable is.
You can divide thru the entire fraction by the variable with the largest degree so n
cubed is the highest degree variable here in our fraction.
If we had an n to the 4th, we would divide thru by n to the 4th.
But if we divide thru by n cubed, what we’ll get is just 10 in the numerator because 10n
cubed divided by n cubed is just 10, the n cubed cancels.
And then when we divide through the denominator here, n cubed divided by n cubed will give
us 1 and then 1 divided by n cubed will stay right here.
So we divided each term in our fraction in both the numerator and denominator by n cubed.
The reason we do this is because now when we evaluate an infinity, we know that when
the denominator of a fraction becomes very, very, very large as
it will when we start plugging in very large numbers, this whole fraction here is going
to become zero.
No matter what’s in the numerator unless it’s also infinity, the denominator will
be infinitely large and that fraction will become zero, which means that what we’re
left with here as n approaches infinity is just 10/1 which of course is just 10.
So 10 is going to be L. Remember here from our theorem, we were looking for L. And L
needed to be greater than zero.
We needed to make sure that L is greater than zero and in this case, it is.
10 is obviously greater than zero.
We have therefore met all conditions that we need to in order to be able to use the
limit comparison test.
So now that we know for a fact that we can use the limit comparison test, we need to
compare b sub n and a sub n.
So be sub n, if it converges or diverges, then a sub n is going to do the same thing.
So we know that b sub n is the harmonic series which we know diverges which means that a
sub n must also diverge.
So we can say a sub n diverges by limit comparison test.
And that’s it.
That’s our final answer.
I hope that video helped you guys and I will see you in the next one.
Bye!