If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains ***.kastatic.org** and ***.kasandbox.org** are unblocked.

Main content

Current time:0:00Total duration:2:57

AP.CALC:

LIM‑7 (EU)

, LIM‑7.A (LO)

, LIM‑7.A.10 (EK)

what are all positive values of P such that the series converges so let's see we have the sum from N equals 1 to infinity of negative 1 to the n plus 1 times P over 6 to the N so there's a couple of things that might jump out at you this negative 1 to the n plus 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 and actually let's just write it out this is going to be see what N equals 1 this is going to be to the second power so it's going to be positive 1 10 so it's going to be P over 6 and then when n is 2 this is going to be to the third power so it's going to be minus P over 6 squared then plus P over 6 to the third power and I could even write to the first power right over here then minus P over 6 to the fourth power and we're going to just keep going plus/minus on and on and on and on forever so this is clearly this is a classic alternating series right over here and so we can actually apply our alternating series test and 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 monotonic alee 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 0 so the limit as n approaches infinity of P over 6 to the nth power also has to be equal to 0 so under what conditions is that going to be true well to meet either one of those conditions P over 6 has to be less than 1 if P over 6 was equal to 1 for example P was 6 well then we wouldn't be monotonically decreasing every term here would just be 1 it would be 1 to the 1 1 squared on and on and on and if P is greater than 6 well then every time we multiply by P over six 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 P over six needs to be less than one and so multiply both sides by six and you get P it needs to be less than six and they told us for what are what are all the positive positive values of P so we also know that P has to be greater than zero so P is greater than zero and less than six which is that choice right over here and 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 one to the N and so we're just going to have this would be 1 this would be 1 would be 1 minus 1 plus 1 and on and on and on forever so definitely like that first choice

AP® is a registered trademark of the College Board, which has not reviewed this resource.