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Current time:0:00Total duration:4:50

AP.CALC:

LIM‑7 (EU)

, LIM‑7.A (LO)

, LIM‑7.A.9 (EK)

- [Instructor] So we're
given a series here, and they say, "What series should we use "in the limit comparison test?" Let me underline that,
"The limit comparison test, "in order to determine
whether S converges?" So let's just remind ourselves about the limit comparison test. If we say, if we say
that we have two series, and I'll just use this notation, a sub-n, and then another series, b sub-n. And we know that a sub-n
and b sub-n are greater than or equal to zero for all n,
for all n, if we know this, then if, then if the limit
as n approaches infinity of a sub-n over b sub-n is equal
to some positive constant. So zero is less than that
constant is less than infinity, then either both converge,
then both converge, both converge, both converge, or both diverge, both diverge. And it really makes a lot of
sense, because it's saying, look, as we get into our
really large values of n, as we go really far out
there in terms of the terms, if our behavior starts to look
the same, then it makes sense that both these series
would converge or diverge, and we have an introductory
vizio, (chuckles) we have an introductory video
on this in another video. So let's think about what,
if we say that this is our a sub-n, if we say that this
is a sub-n right over here, what is a series that we
can really compare to? That seems to have the same
behavior as n gets really large? Well this one just
seems to gets unbounded. This one doesn't look that similar. Has a three to the n minus
one in the denominator, but the numerator doesn't behave the same. This one over here is interesting because we could write this, the is the same as a sum
n equals one to infinity. We could write this is two
to the n over three to the n, and these are very similar. The only difference between
this and this is that in the denominator here, or
in the denominator up here we have a minus one, and down here we don't have that minus one, and so it makes sense given
that that's just a constant, that as n gets very large, that
these might behave the same. So let's try it out. Let's find the limit, and we also know that the a
sub-n's and the b sub-n's, if we say that this over here is b sub-n, we say that's b sub-n, that
this is going to be positive, or this is going to be
greater than or equal to zero for n equals one, two,
three, so for any values this is going to be greater
than or equal to zero, and the same thing right over here. It's gonna be greater
than or equal to zero for all of the n's that we care about. So we meet these first
constraints, and so let's find the limit as n approaches
infinity of a sub-n, which is, I'll write in that red color, which is two to the nth power
over three to the n minus one over b sub-n, over two to the
nth over three to the nth. So let me actually do a little algebraic manipulation right over here. This is going to to be the same thing as two to the nth over
three to the n minus one, times three to the n, over two the n. Divide the numerator and the denominators by two to the n, those cancel out, and so this will give
us, this will give us three to the n over
three to the n minus one. We can divide the numerator
and the denominator by three to the n, and that'll give us one over one minus one
over three to the n. So we could say this is
the same thing as the limit as n approaches infinity
of one over one minus one, over three to the n. Well what's this going to be equal to? Well as that approaches
infinity, this thing, one over three to the n,
that's just gonna go to zero. So this is, this whole thing
is just going to approach one. And one is clearly
between zero and infinity, so the destinies of these
two series are tied. They either both converge
or they both diverge. So this is a good one to use
a limit comparison test with, and so let's think about it. Do they either both converge,
or do they both diverge? Well this is a geometric
series, our common ratio here is less than one, so this
is going to converge, this is going to converge. And because this one converges, by the limit comparison test, our
original series S converges. Converges, and we are done.

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