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### Course: Integral Calculus > Unit 5

Lesson 6: Comparison tests# Direct comparison test

If every term in one series is less than the corresponding term in some convergent series, it must converge as well. This notion is at the basis of the direct convergence test. Learn more about it here.

## Want to join the conversation?

- I don't understand why the sequence can't oscillate just because its terms are all non-negative.

Can someone please explain why? I mean, for example, can't a sequence be like 3,4,3,4,3,4,3,4,3,4,3,4,... which is oscillating between two values but has all non-negative values??(37 votes)- From the author:In order for the sequence of partial sums to oscillate, you would need negative terms in the series. For example, to go from 4 to 3, you'd have to add -1. We made the assumption that every term in the sum is non-negative.(56 votes)

- 3:10, I don't quite understand how we can define that
`Σ a_n`

is "smaller" than`Σ b_n`

? We compare the summations between them or what?(9 votes)- One of the assumptions established at the beginning is that each term in the "a" sequence is less than or equal to the corresponding term in the "b" sequence (that is, a_n is less than or equal to b_n). As a result, the sum of the terms in the "a" sequence has to be less than or equal to the sum of the terms in the "b" sequence.(8 votes)

- Hello there,

If one series is convergent, the another is also convergent.

If one series is divergent, the other one is also divergent.

or

If one greater series is convergent, the another is also convergent.

If one smaller series is divergent, the other one is also divergent.

Which one do you want to mean in the video? I don't need to care about which is bigger or smaller, do I?

Thanks

Aung(4 votes)- The second version you said is correct.

We say a series diverges if it adds all the way to infinity, right? It makes sense that if there's a series that diverges, a series larger than that one will also diverge. This happens because the smaller one went to infinity, so of course the bigger one will too. Similar thinking goes for convergent series. Convergence is defined as a series that adds up to a finite number. So if there's a series smaller than one that converges to a finite number, it wouldn't make sense for the smaller one to all the way up to infinity because a series larger than it converged.

Hope that helped.(5 votes)

- why does a_n,b_n have to be >= 0? Why can not both be nagative, just not equal to 0?(5 votes)
- what if b(n) diverges , will that make a(n) diverges.

thanks(3 votes)- Since b(n) is ≥ a(n) for all n, if b(n) diverges, it says nothing about a(n) and the divergence test is not applicable.

This is what you must understand about the divergence test.....

If you have two different series, and one is ALWAYS smaller than the other, THEN

1) IF the smaller series diverges, THEN the larger series MUST ALSO diverge.

2) IF the larger series converges, THEN the smaller series MUST ALSO converge.

You should rewatch the video and spend some time thinking why this MUST be so. Understanding this is paramount to moving forward with respect to understanding the divergence and convergence of series.

Keep Studying and Keep Asking Questions!(4 votes)

- At0:38, Sal writes a[n],b[n] <=0.

Is this just a given part of the problem i.e. is he saying "Imagine two series, a[n] and b[n] where both of them are >=0..."

OR, did Sal use some information about the series to determine that they are nonnegative?(2 votes)- Sal is defining the properties of a[n] and b[n] so that the comparison test is valid and so that you can see the what is required of the series you wish to test in order to apply the "Comparison Test" - so your first thought is correct although this isn't a problem, it is (the basis of) a theorem.(3 votes)

- We always skip the case when the sequence contains negative numbers.

So what test (or method) can be applied when we examine a sequence with negative numbers??

for example,

a sequence like,

1, -0.1, 0.01, -0.001, 0.0001.........

its series converge to 0.9090909090..................a repeating decimal.

do we have a rule for them?(1 vote)- The Direct Comparison test only applies when the terms in both series are nonnegative. Then, it does not apply in your case. However, the Alternating series test does apply. This is discussed further on in the list of videos. The web address is:

https://www.khanacademy.org/math/calculus-home/series-calc/ratio-and-alternating-series-tests-calc/v/alternating-series-test

You have realized that the sequence forms a geometric series when the terms are summed together, and that the rules governing a geometric series apply.(4 votes)

- What's the intuition behind why if the smaller series diverges then the larger series diverges? Why not if the larger series diverges then the smaller series diverges too? And I have the same questions except regarding convergence instead of divergence.(1 vote)
- If 𝑥 > 𝑦 and 𝑦 > 𝑧, it follows that 𝑥 > 𝑧. So if we consider divergence to be approaching ∞, then if a series "diverges", and the partial sums of another series are larger than the original divergent series, then it follows that larger series must also be divergent.

Using our previous inequalities, if we are given that the infinite series 𝑦 + 𝑦 + 𝑦 + ... diverges, then 𝑥 + 𝑥 + 𝑥 + ... also diverges since it is "larger than something that is already shown to be infinite" loosely speaking.

Comment if you have questions!(2 votes)

- why a_n and b_n terms must be positive values not negative? if they negative terms, they are going to oscillate in all cases ?(1 vote)
- So there is possibility the series can go to negative infinity if you remove the constraint.

Remember the questions you have asked regarding the converging monotone theorem. This is analogous to that.(2 votes)

- so does both laws can go viceverse?

i mean if the greater one diverges does it mean that the smaller one also diverges or if the smaller one converges does it mean that the greater one converges?

and if not so why did the test go that way?(1 vote)- nope, think of this like 2 ordinary cups, one can fit inside of the other and this is like saying the inner cup can contain 100ml of water and the outer cup can contain only 50ml. it is impossible but the reverse is perfectly normal.(1 vote)

## Video transcript

- [Voiceover] So let's
get a basic understanding of the comparison test when
we are trying to decide whether a series is
converging or diverging. So let's think of two series. So let's say that I have
this magenta series here. It's an infinite series from n equals one to infinity of a sub n. We're speaking in generalities here, and let's have another one. That's the series b sub n
from n equals one to infinity, and we know some things
about these series. The first thing we know
is that all the terms in these series are non-negative. So a sub n and b sub n
are greater than or equal to zero, which tells us
that these are either going to diverge to positive infinity or they're going to converge
to some finite value. They're not going to
oscillate, because you're not going to have negative values here. You can't go to negative
infinity, because you don't have negative values here. Now, let's say we also know that each of the corresponding
terms in the first series are less than or equal
to the corresponding term in the second series. Less than or equal to b sub n. Once again, this is true for all the ns that we care about. So n equals one, two, three, all the way on, and on, and on. So the comparison test
tells us that because all the corresponding terms of this series are less
than the corresponding terms here, but they're greater than zero, that if this series converges,
the one that's larger, if this one converges, well then the one that is smaller than
it, or I guess when we think about it is kind
of bounded by this one, must also converge. I'm not doing a formal
proof here, but hopefully that gives you a little bit of intuition. So the comparison test tells
us if, I guess in my brain the larger series, the one
whose corresponding terms are at least as large as
the ones here, if this one converges, if this one
doesn't go unbounded towards infinity, it sums
to some finite value, then that tells us that the
one that is in some ways I guess you could say
smaller must also converge. So this one right over
here must also converge. So that must also converge. So why is that useful? We'll see this in future videos. Well if you find if you're
looking you have your a sub n and you're like
gee, I wish I could prove that it converges, I kind
of have a gut feeling it converges, the comparison
test tells us, well, just find another series
that whose corresponding terms are at least as
large as the corresponding terms here, and if you can
prove that one converges, then you're good with this one. Of course it would only apply to the case where your original
series, each of the terms are non-negative. Now what if you went the other way around? What if you could prove
that the magenta series, the smaller one, and I
guess I could kind of put them in quotes,
this one right over here is the smaller, I guess
each of its corresponding terms are smaller, what if you could prove this one diverges? Well if this one
diverges, it's going to go unbounded to infinity. It's not going to go to negative infinity. All the terms are positive. It's not going to diverge because it oscillates between two values. Once again, if it's
oscillating between values the only way you could do that is if you had negative terms here,
so this would kind of be unbounded towards infinity. Well if this one is
unbounded, each of these corresponding terms
are larger, so this one must also be unbounded. So let's write that down. So the comparison test
tells us if our smaller series diverges, if this one diverges, then the larger one must also diverge. So once again, if you wanted to prove that this thing right over
here is going to diverge, and if you have a, once
again you know that all the b sub ns are greater
than or equal to zero and you want to prove it diverges, well, maybe you could try
to find another series where each of the
corresponding terms are less than the corresponding
terms here and you could prove this one diverges,
then you would be all set. We're going to start doing
that in the next few videos.