Welcome back, everyone. So you may recall a time when we learned about sequences. If you're rusty on this, a sequence is very similar to a function where you have an input and then an output, except rather than having just a random input and then getting a random output, a sequence is going to have discrete inputs like 1, 2, 3, 4, and you're going to get an entire list of outputs, which will oftentimes have a pattern. Now what we're going to learn in this video is a way that you can apply these limits where your input approaches infinity to sequences. Basically, your input n is going to get really big. If that sounds scary, don't sweat it because it turns out these rules that we've learned for functions where the limit approaches infinity also apply to sequences. Let's just go ahead and jump into an example to see what these types of problems look like.

We'll start with example a. We're asked to find the sequence here. We have an = 1n. We're asked to find the limit as n approaches infinity for this sequence. To solve this problem, there's something that I notice. Notice that our input is n, and this value is approaching infinity. Because we can see n here is on the bottom of a fraction and doesn't show up on the top, what we know is that this entire fraction is going to go to 0. That's something that we've learned about limits that approach infinity. So this right here would be the answer to our sequence, and that's it. That's the solution right there.

Now if you didn't know to do this trick right here, or you simply wanted to see why this was the case, another strategy you can use if all else fails is to draw out a table. So one of the things I could do is figure out what, say, a1 is. And a1 would be 11, which is just 1, and I could try a10. a10 would be 0.1 or a100. This would be 0.01, and this would just keep going on. Now as we can see, with these inputs, as our number gets bigger and bigger, approaching infinity, notice that our number is getting closer and closer to 0. So we can see that this was a correct calculation on our part.

Now how exactly could we describe this answer that we got? Well, when you have situations where the limit actually exists and you're dealing with the sequence as your input approaches infinity, what you do is you say the sequence converges. In this example, we can see that we did get an answer. We got 0. So, we would say that this sequence converges, and I think that makes sense because I think of it like this: If you have something that's going to infinity, but you're managing to get your output that converges to one answer, you would say that that answer exists because it exists at one point, so it converges.

Now likewise, if the answer does not exist, we would say that the sequence diverges. Let's go ahead and try a couple more examples to see if we can identify these cases. We'll now move to example b. In example b, I have this sequence right here. Now the way that I'm going to solve this is by trying to simplify things. So what I can see is this n here is going to cancel with that one. So we could rewrite this as an = 8 ∙ ( n + 1 )3. What I can see here is that we have an n to the first power on top, and we don't have any n's on the bottom. That means that this entire fraction is going to blow up as n approaches infinity. So what we'd say is that our sequence actually does not exist. And whenever your sequence does not exist, we would say that our answer diverges, our sequence does. So that's the solution. As you can see, it's pretty straightforward. It's just like what we learned about functions.

Now you could have drawn out a table like we did over here, but again, we knew the shortcuts, we were able to solve this. Now let's go ahead and try example c. For example c, we have an = −1n. This is a little bit of a tricky situation because notice we don't have the typical rational function that we did in these other two examples. So how could we solve this one? Well, remember, if all else fails, you can just draw a table. So let's go ahead and do that over here. I'm going to start with a1. Well, a1 would be negative one to the one power, which is just negative one. What if I had a2? Well, that'd be negative one squared, which is positive one. What if I had a3? Well, now we have an odd number of negative ones being multiplied, so we would say that that's just negative again. What about a100? Well, that's an even number of negative ones, so now we have positive one. But a101, that's an odd number, so that would give us negative one. This is kinda weird. We're getting values between negative one and one, and it's just repeating between these values, we're not really seeing a specific answer, and recall that we've actually dealt with this in the past. We've seen in the past that for the function, the sine of x, when x approaches infinity, this function oscillates between 1 and negative one. It just keeps going up and down, and we're seeing that same pattern happen for this sequence. And we know for these types of situations, we say that our sequence or our solution does not exist. And because our sequence here does not exist, we were constantly bouncing back and forth between negative one and positive one, we would say that our sequence actually diverges. So this right here would be the solution to this problem, and that is how you can solve these types of situations where you have sequences. Hope you found this video helpful and let's move on.