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Convergent and divergent sequences

AP.CALC:
LIM‑7 (EU)
,
LIM‑7.A (LO)
,
LIM‑7.A.1 (EK)
,
LIM‑7.A.2 (EK)

Video transcript

Let's say I've got a sequence. It starts at 1, then let's say it goes to negative 1/2. Then it goes to positive 1/3. Then it goes to negative 1/4. Then it goes to positive 1/5. And it just keeps going on and on and on like this. And we could graph it. Let me draw our vertical axis. So I'll graph this as our y-axis. And I'm going to graph y is equal to a sub n. And let's make this our horizontal axis where I'm going to plot our n's. So this right over here is our n's. And this is, let's say this right over here is positive 1. This right over here is negative 1. This would be negative 1/2. This would be positive 1/2. And I'm not drawing the vertical and horizontal axes at the same scale, just so that we can kind of visualize this properly. But let's say this is 1, 2, 3, 4, 5, and I could keep going on and on and on. So we see here that when n is equal to 1, a sub n is equal to 1. So this is right over there. So when n is equal to 1, a sub n is equal to 1. So this is y is equal to a sub n. Now, when n is equal to 2, we have a sub n is equal to negative 1/2. When n is equal to 3, a sub n is equal to 1/3, which is right about there. When n is equal to 4, a sub n is equal to negative 1/4, which is right about there. And then when n is equal to 5, a sub n is equal to positive 1/5, which is maybe right over there. And we keep going on and on and on. So you see the points, they kind of jump around, but they seem to be getting closer and closer and closer to 0. Which would make us ask a very natural question-- what happens to a sub n as n approaches infinity? Or another way of saying that is, what is the limit-- let me do this in a new color-- of a sub n as n approaches infinity? Well, let's think about if we can define a sub n explicitly. So we can define this sequence as a sub n where n starts at 1 and goes to infinity with a sub n equaling-- what does it equal? Well, if we ignore sign for a second, it looks like it's just 1 over n. But then we seem like we oscillate in signs. We start with a positive, then a negative, positive, negative. So we could multiply this times negative 1 to the-- let's see. If we multiply it times negative 1 to the n, then this one would be negative and this would be positive. But we don't want it that way. We want the first term to be positive. So we say negative 1 to the n plus 1 power. And you can verify this works. When n is equal to 1, you have 1 times negative 1 squared, which is just 1, and it'll work for all the rest. So we could write this as equaling negative 1 to the n plus 1 power over n. And so asking what the limit of a sub n as n approaches infinity is equivalent to asking what is the limit of negative 1 to the n plus 1 power over n as n approaches infinity is going to be equal to? Remember, a sub n, this is just a function of n. It's a function where we're limited right over here to positive integers as our domain. But this is still just a limit as something approaches infinity. And I haven't rigorously defined it yet, but you can conceptualize what's going on here. As n approaches infinity, the numerator is going to oscillate between positive and negative 1, but this denominator is just going to get bigger and bigger and bigger and bigger. So we're going to get really, really, really, really small numbers. And so this thing right over here is going to approach 0. Now, I have not proved this to you yet. I'm just claiming that this is true. But if this is true-- so let me write this down. If true, if the limit of a sub n as n approaches infinity is 0, then we can say that a sub n converges to 0. That's another way of saying this right over here. If it didn't, if the limit as n approaches infinity didn't go to some value right here-- and I haven't rigorously defined how we define a limit-- but if this was not true, if we could not set some limit-- it doesn't have to be equal to 0. As long as it-- if this was not equal to some number, then we would say that a sub n diverges.