The comparison test, again.
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Well, here's something that happens.
Well, maybe you're analyzing a couple series, and
the first thing that you do is try to
apply the limit test, and you find that both
cases, the limit of the nth term is zero.
So, at least these series aren't diverging for an obvious
reason, like the limit of the nth term is non zero.
So you got two series that may or may not converge.
But, even though the limit
of a sub n is zero, and the limit of b sub n is zero, it could be that the limit as n
goes to infinity of a sub n over b sub n might be some number L, which is positive.
What does that really mean?
Well one way to think about that is that it's sort of saying something like this.
It's saying that a sub n, b sub n are almost multiples of
each other, like a sub n is almost a multiple of b sub n,
at least when n is really big. I can be more precise with epsilons.
So let's set epsilon equal to L in this case.
Epsilon's going to be a positive number, but
I'm assuming that my limit, L, is positive.
So let's set epsilon equal to L. Then the definition of limit says what?
It says that there's some big N, so that whenever little
n is greater than or equal to big n the distance
between the thing I'm taking the limit of, a sub n
over b sub n, and my limit, L, is less than epsilon.
And in this case, right, epsilon is L.
That lets me compare a sub n and b sub n. Well, how so?
Let me make this assumption that the a sub n's and the b sub n's are non-negative.
I'm going to want that because I'm going to apply the comparison test in a moment.
So I can simplify this a bit.
Instead of making this claim, I can just get
rid of the absolute value bars, it's still true, right.
A sub n over b sub n, minus L is less than L.
You can add L to both sides of that inequality and I get
this, that a sub n over b sub n, is less than 2L.
And now I can multiply both sides by b sub n, and that's okay,
because b sub n is non-negative, so
it doesn't change the direction of this inequality.
And that
tells me that at least for large values of little n, a sub n is less than 2L b sub n.
Now all these pieces are really setting up a comparison test.
How is that going to work?
I got to remember this is only true for large N, but that's okay.
Well suppose that I knew that this series, the sum little
n goes from 1 to infinity of b sub n converged,
well then I would know that this series, the sum little n
goes from 1 to infinity of 2L b sub n also converges.
Right, I can multiply a convergent series just by some number.
But now I'm in the position to apply the comparison test.
Granted this statement's only true for large values of little n but that's
okay, right, because convergence only depends on a tail, so this statement
then implies that this series, the sum of the a sub n's converges, because
this is bigger than a sub n, I mean at least for large values of little n.
So I'm getting a theorem that's telling me if I've got two
series of non-negative terms and this series
converges, then this series converges, provided that
this is true. Let me summarize that.
So, if I've got a sub n's are all non-negative, b sub n's are all
non-negative, the sum of the b sub n's converges, and this limit
statement that the limit of the ratios between the a sub n's and
the b sub n is some finite value L, which is positive, then I
can conclude that this series, the sum of the a sub n's, n goes from 1 to
infinity, converges as well. Now let me exchange the roles of a and b.
So I'd like to be able to start with
the assumption that the series of the a sub n's
converges and then conclude that the series of the b
sub n's converges, but actually that's the exact same statement.
All right, watch this. If I just replace this limit with this
limit, now I'm in the exact same situation, except now b
sub n's and a sub n's roles are exchanged, all right, and that means that if I know
that this series converges, then I know that this series converges as well.
Now put it all together.
Well, since the convergence of a sub n implies the
convergence of b sub n, and the
convergence of b sub n implies the convergence
of a sub n in the presence
of this limit statement, equivalently, this limit statement.
I can simply this a bit, I can say that
if I've got two sequences of numbers, a sub n and
b sub n, both non negative, and this limit exists
and is equal to some number bigger than zero, then this
series converges, if and only if, this series converges.
The sum of the b sub n's converges if and only if, the sum a sub of n's converges.
This convergence test has a name.
This is called The Limit Comparison Test, and it's
one of the situations when two series, in this case
the sum of the b sub n's and the sum of the a sub n's, share the same fate.
They either both converge or they both diverge.
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