We will be adding to our list of series tests today with looking at what's called the direct comparison test, which I will be abbreviating as DCT, and we will also be looking at the limit comparison test, LCT. Both of these require that you're looking at terms that are positive, greater than zero. This specific one, the direct comparison test, It says that a sub n needs to be smaller or at least equal to b sub n for all our values.
And if you know a series converges and you are going to be able to compare it to a sub n, then a sub n converges. If you pick a series that you know diverges and you compare it, then you'll know that b sub n diverges. Now that all goes back to this statement. So think about it. So if a sub n is really big, then what has to be true for b sub n?
b sub n is even bigger. That gives you that. But if b sub n converges, which means it's approaching a smaller number, then a sub n is even going to be smaller than that because a sub n is less than b sub n. So try to understand the comparisons.
Now, we were going to see in the direct comparison test that we want all a sub n's to be smaller than b sub n. But because the convergence of a series is not dependent really on its first several terms, you can modify the test to only require that this is true eventually. As in at some point that is going to be a true statement.
We're also going to be looking at the limit comparison test. So here you just have to know that all your terms are positive. And then most books just give the limit comparison test as this, that if you compare the limit of the ratio of the two series and you get L, where L is a limit that's finite and positive, then these series...
will agree, meaning they either both converge or diverge. However, a few other books will also state that if you have a limit that equals to zero, then they would both converge. And if the limit is infinity, they would both diverge.
This first case is the one that's used most often. And the great thing is it will work even if you accidentally kind of write the fraction upside down. Whether you put the A on top or the B on top. As long as you get that L is a finite and positive limit, then they're still going to agree on convergence or divergence. However, these two cases are special.
You have to make sure. that the fraction is what I would call right side up. In other words, the a sub n has to be on top of b sub n for this to work.
Now you're going, well, what do I know? a sub n, b sub n. These two series is what you're going to choose to compare.
And how are you going to know what to choose? That's the hard part. We're just going to choose things, and that's why we're going to practice. So, let's do this. I also want to say that just with the direct comparison test, the limit comparison test can be modified to require that the terms be positive at some point, as in eventually.
At the beginning, you may have a few negatives, but as long as they eventually are all positive, the test will still hold. The only way we get better at all of these tests is to practice. So let's just dive in.
So we're going to determine the convergence or divergence. Clearly state your reasons. Now remember, you can use any test that you think might work.
But these first few are designed where you're going to have to use either the direct comparison test or the limit comparison test. Because these are not going to be geometric, they're not telescoping, they're not... A p-series, the nth term test fails, and I just like to avoid the integral test as all possible.
Alright, so let's look at number one, and let's think about that comparison test, meaning you have to pick something that you know, alright? Pick a series that is known. When I say is known, like a geometric series, and you know what r equals to.
A p-series where you know what p equals to. Those are really what you're going to compare it to. How do you know?
Well, let's look at this. Look at the powers of n and kind of disregard the numbers. Wouldn't you think that it would reduce to 1 over n cubed? Do I know anything about the series of 1 over n cubed?
Yes, I do. This is a known series. This is a p-series where p is 3, which is greater than 1. So I know.
this converges. But I now have to compare it. So the direct comparison test.
What do I have to do? Well the first thing I have to do is I have to show an inequality. So if what I'm picking converges then I'm going to say alright 1 over n cubed converges.
What I'm comparing it to has to be smaller. Think about that. If this converges to a value and this is smaller, then it also has to converge. So I have to show that that inequality is true. How can I do that?
Well, one of the easiest ways is to do some cross multiplication, but you have to be very careful and keep it in the right inequality order, meaning this has to go on the left-hand side. So, n cubed times n minus 1. Is that really less than or equal to 1 times n to the fourth plus 2? So let's see.
I have n to the fourth minus n cubed. Is that less than or equal to n to the fourth plus 2? Is that true?
Yes it is because of that negative. So if you take a negative number and cube it, you're still going to get something negative which is definitely less than positive 2. So I can then conclude that the series, the given series, also converges because of the direct comparison test. I know that's weird. Just keep going. All right, so let's look at example number two.
Again, I have to pick something that... that I know about. So I just kind of ignore all the numbers and say, well, wouldn't this kind of look like one over the square root of n?
And you're going, what's that? Well, remember the square root. is the half power. That's also a p-series where p is one-half which is less than one so this diverges.
That's different. That's okay. I know that that diverges. So I'm going to pick my known series which is one over, I'm just going to rewrite that as the square root of n and if that diverges I have to show that that is smaller than my given series because remember I want want it to agree.
So one more time, if this diverges and this one is even bigger, then it would also diverge. So I have to show that inequality. So again, I'm going to cross multiply being very careful.
1 times the square root of n, is that really greater than or equal to 1 times 2 plus the square root of n. Is that true? Well, no. Square root of n is not bigger than 2 plus the square root of n. So this is not true.
Did I do anything wrong? No. I just have to use another test.
Well, since the direct comparison test didn't work, maybe I should try the limit comparison test. Okay, so the direct comparison test failed. So let's try the limit comparison test. The neat thing about the limit comparison test, I'm going to still use the known series. So I'm going to still use all of that information.
So I'm going to compare it to, I'm just going to rewrite this again, the square root of n, where p is 1 half, which is less than 1, which diverges. So the limit comparison test. So I'm going to take the limit as n approaches infinity. What I choose to compare it to is going to go in the denominator.
If you always do that, you really can't go wrong. What I'm comparing it to, or the given, is going to go on top. Well that's yuck. So let's just do some algebra. So remember when you divide by a fraction, you flip and multiply by the reciprocal.
So this would be... This would come up here, and I would have then the square root of n over 2 plus the square root of n. And then can I take the limit? Well, sure, because these square roots are the same power, so I have the same power over the same power, so that limit is just 1, which is good because that is a finite positive number.
And my limit comparison test says if I have a finite positive number as a limit, then these two series agree. So I'm going to say, therefore, the given series diverges because of the limit comparison test. Okay, so I've shown what I've compared it to, how I know what I'm comparing it to diverges, and then... I've taken the limit. I've made a conclusion.
Written all my steps down.