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## Comparison tests for convergence

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# Limit comparison test

AP Calc: LIM‑7 (EU), LIM‑7.A (LO), LIM‑7.A.9 (EK)

## Video transcript

- Let's remind ourselves,
give ourselves a review of the Comparison Test,
see where it can be useful, and maybe see where it might
not be so useful, but luckily we'll also see the Limit Comparison Test, which can be applicable in a
broader category of situations. So we've already seen
this, we want to prove that the infinite series from
n equals one to infinity of one over two to the n plus one
converges how can we do that? Well, each of these terms are
greater than or equal to zero, and we can construct another series where each of the corresponding
terms are greater than each of these corresponding terms. And that other series,
the one that jumps out at, that will likely jump out
for most folks would be one over two to the n,
one over two to the n is greater than, is greater
than, and all we would have to really say is
greater than or equal to, but we could actually explicitly say it's, oh I'll just write it's
greater than or equal to, one over two to the n plus one for, n is equal to one, two, all
the way to infinity - why? Because this denominator
is always going to be greater by one if you're
denominator is greater, the overall expression
is going to be less, and because of that, because
each of these terms are, they're all positive, this
one, each corresponding term is greater than that one, and
by the Comparison Test, because this one converges,
this kinda provides an upper bound, because this series
we already know converges, we can say, so because this one converges, we can say that this one converges. Now, let's see if we can
apply a similar logic to a slightly different series. Let's say we have the series the sum from n equals one to infinity of one over two to the n minus one. In this situation can we do just the straight up Comparison Test? Well no, because you cannot say that one over two to the n is
greater than or equal to one over two to the n minus one. Here, the denominator is lower, means the expression is greater,
means that this can't, each of these terms can't provide
an upper bound on this one. This one is a little bit larger, but on the other hand you're
like, 'Ok, I get that.' But look, as n gets large,
the two to the n is going to really dominate the minus
one or the plus one, or this one has nothing there,
this one just has two to the n. The two to the n is
really going to describe the behavior of what this thing does. And I would agree with you, but we just haven't proven that it actually works, and that's where the Limit
Comparison Test comes in helpful. So let me write that down,
'Limit, Limit Comparison Test, 'Limit Comparison Test',
and I'll write it down a little bit formally,
but then we'll apply it to this infinite series right over here. So what Limit Comparison Test tells us, that if I have two infinite series, so this is going from n equals
k to infinity, of a sub n, I'm not going to prove it here, we'll just learn to apply it first. This one goes from n equals
k to infinity, of b sub n. And we know that each of
the terms, a sub n, are greater than or equal to zero, and we know each of
the terms, b sub n, are actually we're just going
to say, greater than zero, because it's actually
going to show up in the denominator of an expression, so we don't want it to be equal to zero, for all the n's that we care about. So for all n equal to k, k plus one, k plus two, on and on, and on and on, and, and this is the key, this
is where the limit of the Limit Comparison Test comes into play, and, if the limit, the limit
as n approaches infinity, of a sub n over b sub n, b sub n is positive and finite,
is positive and finite, that either both series converge, or both series diverge,
so let me write that. So then, that tells us that either, either both series converge, or both diverge, which
is really really useful. It's kind of a more
formal way of saying that, 'Hey look, as n approaches
infinity, if these have similar 'behaviors then they are either going to 'converge, or they're
both going to diverge.' Let's apply that right over here. Well if we say that our b sub n is one over two to the n, just like we did up there, one over two to the n, so we're going to compare, so these two series right over
here, notice it satisfies all of these constraints,
so let's take the limit, the limit as n approaches infinity of a sub over b sub n, so it's going to be one over two to the n minus one over one over two to the n, and
what's that going to be equal to? Well that's going to be equal to the limit as n approaches infinity, of two, if you divide by
one over two to the n that's just going to be the
same thing as multiplying by two to the n, so it's going to be two to the n over two to the n minus one. Over two to the n minus
one, and this clearly, what's happening in the
numerator and the denominator, these are approaching the
same, well we can even write it like, we can
even write it like this, divide the numerator and
denomiator by two to the n, if you want, although it's probably going to jump out at you at this point. So, limit as n approaches infinity, let me just scroll over
to the right a little bit. If I divide the numerator by two to the n, I'm just going to have one. If I divide the denominator
by two to the n, I'm going to have one minus
one over two to the n. And now it becomes clear,
this thing right over here is just going to go to zero, and you're going to have one over one. The important thing is that this limit is positive and
finite because this thing is so this thing right over
here, is positive and finite, the limit is one is positive and finite, if this thing converges and
this thing converges, or this thing diverges, and this diverges, well, we already know
this thing converges, it's a geometric series where the common ratio is less than
one, and so therefore this must converge as well,
so that converges as well.

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