Worked example: alternating series | Series | AP Calculus BC | Khan Academy

What are all positive values of P such that the series converges?

So let's see, we have the sum from n equal 1 to infinity of ((-1)^{n + 1} \frac{p}{6^{n}}).

There's a couple of things that might jump out at you. This ((-1)^{n + 1}) as (n) goes from 1 to 2 to 3, this is just going to alternate between positive 1, negative 1, positive 1, negative 1. So we're going to have alternating signs, so that might be a little bit of a clue of what's going on.

Actually, let's just write it out. This is going to be

• when (n = 1), this is going to be (1^{2}), so it's going to be positive 1, so it's going to be (\frac{p}{6});
• when (n = 2), this is going to be (1^{3}), so it's going to be minus (\frac{p}{6^{2}});
• then plus (\frac{p}{6^{3}});
• and I could even write (\frac{p}{6^{1}}) right over here;
• then minus (\frac{p}{6^{4}})
• and we're going to just keep going plus minus on and on and on and on forever.

So this is clearly a classic alternating series right over here. We can actually apply our alternating series test. Our alternating series test tells us that if this part of our expression, the part that is not alternating in sign, I guess you could say, if this part of the expression is monotonically decreasing, which is just a fancy way of saying that each successive term is less than the term before it.

And if we also know that the limit of this as (n) approaches infinity, that also has to be equal to zero. So the limit as (n) approaches infinity of (\frac{p}{6^{n}}) also has to be equal to zero.

So under what conditions is that going to be true? Well, to meet either one of those conditions, (\frac{p}{6}) has to be less than 1. If (\frac{p}{6}) was equal to 1, if for example (P) was 6, well then we wouldn't be monotonically decreasing. Every term here would just be one. It would be (1^{1}), (1^{2}), and on and on and on.

And if (p) is greater than 6, well then every time we multiply by (\frac{p}{6}) again we would get a larger number over and over again, and the limit for sure would not be equal to zero.

So we could say (\frac{p}{6}) needs to be less than 1. Multiply both sides by 6 and you get (P) needs to be less than 6.

They told us for what are all the positive values of (P). So we also know that (P) has to be greater than zero. Therefore, (p) is greater than zero and less than six, which is that choice right over here.

Once again, we're not going to say less than or equal to six, because if (P) was equal to six, this term is going to be (1^{n}) and so we're just going to have this. Would be one, this would be one. It would be 1 minus 1 plus 1 and on and on and on forever.

So definitely like that first choice.