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## Calculus, all content (2017 edition)

### Course: Calculus, all content (2017 edition) > Unit 7

Lesson 9: Comparison tests# Worked example: limit comparison test

AP.CALC:

LIM‑7 (EU)

, LIM‑7.A (LO)

, LIM‑7.A.9 (EK)

To use the limit comparison test for a series S₁, we need to find another series S₂ that is similar in structure (so the infinite limit of S₁/S₂ is finite) and whose convergence is already determined. See a worked example of using the test in this video.

## Want to join the conversation?

- Does it matter which we set as "a" and "b" during the limit comparison test?(18 votes)
- Technically, it doesn't matter, because the only requirements for the limit is that it is finite and positive (0 < c < infinity). Because the exact value is not required for this test, it does not matter which is 'a' and which is 'b' for this type of problem.(22 votes)

- Why is it that, even though (2/3)^n is smaller than (2^n)/(3^n)-1, Sal still used it for the comparison? I would have understood if it diverged, because a>b in that case, but since it is converging, isn't b>a supposedly?(6 votes)
- That concept is applied to the direct comparison test. For the limit comparison test, it does not matter if one series is greater than or less than the other series; as long as the limit of their ratios approach a positive finite value, then they are either both convergent or both divergent.(7 votes)

- I really do feel like these videos are fabulous for anyone in Freshman Calculus wanting to further develop their basic intuition a little bit.(8 votes)
- Don't all the functions meet the criteria specified? I realize that the chosen one corresponds the behavior better, but that's rather abstract. But surely the other functions are also zero or greater?(5 votes)
- The limits of the other functions would not equal the limit of the given function and thus they would not be an effective comparison. Thus, the quotient of the limits would not equal a constant between 0 and infinity and the limit comparison test would no longer apply. It therefore could not be stated that both functions either converge of diverge.

Always pick the function that best models the given function. Hope this helps.(3 votes)

- Is there a way to find out what the original S converges to?(4 votes)
- The behaviour of the infinite terms of a series is enough to know whether a series converges or diverges ??(1 vote)
- No, not in general.

But ∑(2∕3)^𝑛 is a geometric series with a common ratio of 2∕3,

which is less than 1, and thereby the series converges.

By the comparison test we can then conclude that

∑2^𝑛∕(3^𝑛 − 1) also converges.(2 votes)

- At0:25to0:29, Sal says that series a sub(n) and b sub(n) are greater than or equal to zero. Just to clarify b sub(n) cannot be equal to zero it is only greater than zero since it's the denominator.(1 vote)
- There's nothing saying that 𝑏(𝑛) can not be equal to 0 for some value/s of 𝑛, because we only care about the limit of 𝑎(𝑛)∕𝑏(𝑛) as 𝑛 approaches infinity, and as the video clearly shows, this limit can exist even if 𝑏(𝑛) tends to 0.(2 votes)

- Why does sal subscript sigma with n? Is this common notation?(1 vote)
- if {sqrt{n+1}-1}/{(n+2)^3-1} is a then what should we take as b?(1 vote)
- If the test fails and gives c<0 can you deduce anything?(1 vote)

## Video transcript

- [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.