Alternating series test | Series | AP Calculus BC | Khan Academy

Let's now expose ourselves to another test of convergence, and that's the alternating series test. I'll explain the alternating series test, and I'll apply it to an actual series while I do it to make the explanation of the alternating series test a little bit more concrete.

So let's say that I have some series, some infinite series. Let's say it goes from n equals k to infinity of ( a_n ). Let's say I can write it as, or I can rewrite ( a_n ). So, let's say ( a_n ) I can write so ( a_n ) is equal to ( (-1)^n b_n ) or ( a_n ) is equal to ( (-1)^{n+1} b_n ), where ( b_n ) is greater than or equal to 0 for all the n's we care about. So for all of these integer n's greater than or equal to k.

If all of these things are true and we know two more things, we know number one the limit as n approaches infinity of ( b_n ) is equal to zero and number two ( b_n ) is a decreasing sequence. That lets us know that the original infinite series, the original infinite series is going to converge.

So this might seem a little bit abstract right now. Let's actually use this with an actual series to make it a little bit more concrete. So, let's say that I had the series, let's say I had the series from n equals 1 to infinity of ( \frac{(-1)^n}{n} ). We could write it out just to make this series a little bit more concrete.

When n is equal to 1, this is going to be ( \frac{(-1)^1}{1} ), actually, let's just make this a little bit more interesting. Let's make this ( (-1)^{n+1} ). So when n is equal to 1, this is going to be ( \frac{(-1)^2}{1} ), which is going to be 1. And then when n is 2, it's going to be ( \frac{(-1)^3}{2} ), which is going to be negative one-half.

So it's minus one-half plus one-third minus one-fourth plus minus and it just keeps going on and on and on forever. Now, can we rewrite this ( a_n ) like this? Well, sure. The ( (-1)^{n+1} ) is actually explicitly called out. We can rewrite our ( a_n ).

So let me do that. So, ( a_n ), which is equal to ( \frac{(-1)^{n+1}}{n} ), this is clearly the same thing as ( (-1)^{n+1} \times \frac{1}{n} ), which we can then say this thing right over here could be our ( b_n ).

So, this right over here is our ( b_n ), and we can verify that our ( b_n ) is going to be greater than or equal to zero for all the n's we care about. So our ( b_n ) is equal to ( \frac{1}{n} ). Now clearly this is going to be greater than or equal to zero for any positive n.

Now, what's the limit as ( b_n ) approaches? What's the limit of ( b_n ) as n approaches infinity? The limit of, let me just write ( \frac{1}{n} ) as n approaches infinity is going to be equal to 0. So we satisfy the first constraint.

And then this is clearly a decreasing sequence. As n increases, the denominators are going to increase, and with a larger denominator, you're going to have a lower value. So we can also say ( \frac{1}{n} ) is a decreasing sequence for the n's that we care about.

So this is satisfied as well. And so based on that, this thing right over here is always greater than or equal to zero. The limit as ( \frac{1}{n} ) or ( b_n ) as n approaches infinity is going to be zero. It's a decreasing sequence. Therefore, we can say that our original series actually converges.

So, the series from n equals 1 to infinity of ( \frac{(-1)^{n+1}}{n} ), and that's kind of interesting because we've already seen that if all of these were positive, if all of these terms were positive we just have the harmonic series, and that one didn't converge, but this one did. Putting these negatives here do the trick.

And actually, we can prove this one over here converges using other techniques, and maybe if we have time actually, in particular, the limit comparison test. I'll just throw that out there in case you are curious.

So this is a pretty powerful tool. It looks a little bit about like that divergence test, but remember the divergence test is really only useful if you want to show something diverges. If the limit of your terms does not approach zero, then you say, okay, that thing is going to diverge.

This thing is useful because you can actually prove convergence. Now once again, if something does not pass the alternating series test, that does not necessarily mean that it diverges. It just means that you couldn't use the alternating series test to prove that it converges.