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# Convergent and divergent sequences

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.