If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains ***.kastatic.org** and ***.kasandbox.org** are unblocked.

Main content

Current time:0:00Total duration:8:59

AP Calc: LIM‑7 (EU), LIM‑7.A (LO), LIM‑7.A.11 (EK)

- [Voiceover] We already have a lot of experience with the geometric series. For example, if I have the
infinite geometric series starting at N equals K to
infinity of R to the N, which would be R to the K
plus R to the K plus one plus R to the K plus two and keep going on and on and on forever. There are a few things we've
already thought about here. We know what the common ratio is. So the common ratio,
which is the ratio between consecutive terms, is going
to be R to the K plus one. Actually, let me just write
this as R to the N plus one. I don't want to fixate on
this K right over here. So R to the N plus one over R to the N. Well, this is just going to be equal to R. Anything to the N plus one over
that same thing with the N, that's just going to be equal
to R, or R to the first power, and you see that here. When you go from one term to another you are just multiplying, you
are just multiplying by R, and this is all review. If it is not review, I encourage you to watch the videos on geometric series. And what's interesting
about this is we've proven to ourselves, in the videos
about geometric series, that if the common ratio,
if the absolute value of the common ratio is less than one, then the series converges. And if the absolute value of R is greater than or equal to one, then the series diverges. And that makes sense,
we've proven it as well, but it also makes
logical sense that, look, if that absolute value
of R is less than one then each term here is going
to go down by that common, or it's going to be multiplied
by that common ratio, and it's going to decrease
on and on and on and on until it makes sense
that, even though this is an infinite sum, it will
converge to a finite value. Now, with that out of the way for review, let's tackle something a
little bit more interesting. Let's say that we want to look at a, we want to figure out
whether a series like this, so starting at N equals five to infinity of, let's say, N to the tenth power. The numerator is growing quickly. N to the tenth power over N factorial, and factorial we know also
grows very, very quickly, probably it grows much faster than even a high degree polynomial like this, or a high degree term like this, but how do we prove that it converges? We could definitely the diverges test to show that this does not diverge, but how do we prove that
this actually converges? And maybe we can use a little
bit of our intuition here. Well, let's see if we can
come up with a common ratio. So let's see if there's
a common ratio here. So let's take the N plus oneth term, which is going to be N
plus one to the tenth power over N plus one factorial and
divide that by the Nth term, so let's divide that by N to
the tenth over N factorial. Well, dividing by a fraction,
or dividing by anything, is equivalent to multiplying
by it's reciprocal, So let's just multiply by the reciprocal, so times N factorial over N to the tenth. Remember all I'm trying to do
is exactly right over here. See if there is some
type of a common ratio. Well, if we do a little algebra here N plus one factorial, and
factorial algebra is always fun. This is the same thing as N
plus one times N factorial. Times N factorial. You're not use to seeing order of operations with factorials, but this factorial only applies
to this N right over here, and why is that useful? Because this N factorial
cancels with this N factorial, and we're left with N plus
one to the tenth power over N plus one times N to the tenth. Times N to the tenth. So I know what you're thinking. "Hey, wait, look, this isn't
a fixed common ratio here." The ratio between consecutive terms here, when I took the N plus oneth
term divided by the Nth term, it's changing depending on what my N is. This ratio, I guess you
could say, is a function of N so this doesn't seem too useful, but what if I were to say, "O.K., well look, with any of these series "we really care about
the behavior as our N's "get really, really, really large as the limit, as our N's, go to infinity." So what if we were to
look at the behavior here, and if this is approaching some actual values as N approaches infinity, well, it would make conceptual sense that we could kind of think of that as the limit of our common ratio. So let's do that. Let's take the limit as N
approaches infinity of this thing. So remember what we're doing here. This isn't just kind of some voodoo. Conceptually, let me copy and paste this. Conceptually, we're just trying to think about what is the common ratio approach? What is the ratio between
consecutive terms? The N plus one term and the Nth term as N gets really, really,
really, really large? And what we see here is
we have this up here, we don't even have to multiply it out, this is going to be N to the tenth plus a bunch of other stuff. It's going to be a
tenth degree polynomial, and this down here is going to be, actually you can figure it out, this is going to be N to the
11th plus N to the tenth power, so the limit as N approaches infinity, well this denominator is going
to grow faster than that. You could divide the numerator and the denominator by N to the tenth, actually the numerator and the
denominator by N to the 11th, and everything up here
is going to go to zero, and this is going to be one, and so the limit, right over here, this limit is going to be equal to... This limit is going to be equal to zero. As N approaches infinity one way the ratio between consecutive terms approaches zero. So this seems, if we use
the logic from our common ratio up here when we
looked at geometric series, this is clearly not a geometric series, but we would say, "Well, look, as N gets really, really, really
large the ratio between "consecutive terms gets smaller
and smaller and smaller. "Hey, maybe we could
make the same conclusion. "Maybe we could make
the same conclusion that "therefore this series
actually converges." The series actually converges,
so N factorial converges. So can we make this statement? And the answer is yes we can, and what allows us to
do it is the ratio test. Is the ratio test. Let me write that down. The ratio test. And all the ratio test tells us is if we have an infinite series... If we have an infinite series, and I can say from N equals K to infinity, and if we take the limit
as N approaches infinity, of the absolute value
of A to the N plus one over A to the N, and over here I didn't take the absolute value of that limit, but we absolutely could have taken the absolute value of that limit if all of these terms right
over here were positive, so the absolute value would
be the same thing as itself. If this approaches some limit, if this approaches some limit, so once again, this limit is
what is going to be our ratio between consecutive terms
as N gets larger and larger, the ratio tells us if L is less than one, which was the situation here,
L is clearly less than one, then the series converges. Some definitions will say
it's absolutely convergent, it converges absolutely,
the absolute value of the series converges,
but then that also means that the series itself converges. If L is greater than one
then series diverges, Series diverges. and if L is equal to one it's
inconclusive, we don't know. We would have to do some other test to prove whether it converges or diverges. So that's the essence of the ratio test. Find the absolute value of the ratio between consecutive terms, take the limit as N approaches infinity, if that approaches an actual limit, and that limit is less than
one, then the series converges, and it's really based on
the same fundamental idea that we saw with the common
ratio of geometric series.