Worked example: limit comparison test | Series | AP Calculus BC | Khan Academy

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_n ) and then another series ( b_n ), and we know that ( a_n ) and ( b_n ) are greater than or equal to zero for all ( n ). If we know this, then if the limit as ( n ) approaches infinity of ( \frac{a_n}{b_n} ) is equal to some positive constant (so ( 0 < c < \infty )), then either both converge or both diverge.

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. We have an introductory video on this in another video.

So let's think about, what if we say that this is our ( a_n )? If we say that this is ( a_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 seems to get unbounded. This one doesn't look that similar; it has a ( 3^{n-1} ) in the denominator, but the numerator doesn't behave the same.

This one over here is interesting because we could write this. This is the same thing as ( \sum_{n=1}^{\infty} ) we could write this as ( \frac{2^n}{3^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. So it makes sense, given that that's just a constant, that as ( n ) gets very large, these might behave the same.

So let's try it out. Let's find the limit. We also know that the ( a_n ) and ( b_n )--if we say that this right over here is ( b_n ), if we say that's ( b_n ), that this is going to be positive or this is going to be greater than or equal to zero for ( n = 1, 2, 3 ). So for any values, this is going to be greater than or equal to zero, and the same thing right over here; it's going to be greater than or equal to zero for all the ( n ) that we care about.

So we meet these first constraints, and so let's find the limit as ( n ) approaches infinity of ( a_n ), which is written in that red color: ( \frac{2^n}{3^{n-1}} ) over ( b_n ) over ( \frac{2^n}{3^n} ).

So let me actually do a little algebraic manipulation right over here. This is going to be the same thing as ( \frac{2^n}{3^{n-1}} \cdot \frac{3^n}{2^n} ). Divide the numerator and denominators by ( 2^n ); those cancel out. So this will give us ( \frac{3^n}{3^{n-1}} ).

Like we can divide the numerator and denominator by ( 3^n ), and that will give us ( \frac{1}{1 - \frac{1}{3^n}} ). So we could say this is the same thing as the limit as ( n ) approaches infinity of ( \frac{1}{1 - \frac{1}{3^n}} ).

Well, what's this going to be equal to? Well, as ( n ) approaches infinity, this thing ( \frac{1}{3^n} ) is going to go to zero. So this whole thing is just going to approach one. One is clearly between zero and infinity, so the destinies of these two series are tied. They either both converge or they both diverge, and so this is a good one to use the 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.

And we are done.