 In this video, we're going to discuss the solution for question 10 on the practice final exam for Math 12-20, in which case we're asked to find the interval of convergence for the power series, where we take the sum where n equals 1 to infinity of x to the n over n times 2 to the n, as you can see on the screen right here. So when it comes to finding the interval of convergence, the first thing we want to find out is the radius of convergence. So for what radius does this thing converge across? In which case, I would recommend using the ratio test unless there's a good reason to use a different convergence test, maybe like the root test. The root test would work okay right here. I do think the ratio test will be a little bit of cleaver of a calculation. So remember with the ratio test, you have to look at the sequence of ratios, hence the name of an plus 1 over an, where we're looking at the sequence which we're adding together, and that includes the power of x right there. And so if you look at an plus 1, this will become x to the n plus 1, we replace each of the n's with the n plus 1, and then at the denominator you get n plus 1, that n became an n plus 1, and then the power of 2 becomes a 2n plus 1. This right here is our, just to label it here, this is our an plus 1 term. The next thing we want to label is the 1 over an term, which would look like, just, it'll just look like the original c, the terms and the sum, just written upside down, taking it to reciprocal. So we get n times 2 to the n over x to the n, and this is all inside the absolute value. My recommendation next when trying to work out this limit associated to the ratio test is put together similar type things like exponentials, the x's, the n's, put these things together. And so, for example, this is a rational expression, n over n plus 1, those are good friends. There's going to be this exponential with regard to 2, so 2 to the n over 2 to the n plus 1. And then finally, there's going to be this x to the n plus 1 sitting above x to the n. And you'll notice I dropped the absolute values with the exception of x right here, because as n goes from 1 to infinity, the n is always going to be positive, so n over n plus 1 is a positive quantity, and exponentials are always positive 2 to the n right here. And so, taking the absolute value of positive terms that are known to be positive is just redundant, but as x could be a negative value, the absolute value is still necessary there. So let's do a little bit of simplification here. When you take something like 2 to the n plus 1, this does factor as 2 to the n times 2 to the first, just by exponent rules. The 2 to the n's would cancel, leaving you just a 2 in the denominator. The same thing's going to happen with x to the n. You'll be left with just an x to the first. And so this thing would simplify as we get the absolute value of x over 2, and then we're going to times that by n over n plus 1. And so that's the amount of algebraic simplification we can do. Now as we take the limit as n goes to infinity, in that situation, the n over n plus 1 as this is just a balanced rational function, this will just approach 1. So we get the absolute value of x over 2 times 1. That is the absolute value of x over 2. Now to be convergent, the ratio test expects that this thing is supposed to be less than 1. And so if you have the absolute value of x over 2 is less than 1, if you solve for the absolute value of x, you'll get the absolute value of x is less than 2. And this right here is our radius of convergence. As some notice we now can answer our question, the radius of convergence is equal to 2 for this exercise. Now that's just the first part of this exercise. It's one of the most important things, we have to identify the radius of convergence. And so this tells us that will be convergent as long as x is less than 2 and greater than negative 2. But what happens at 2, what happens at negative 2? We actually have to plug those values in individually and see what happens. So if we take the first one, x equals 2, if you plug it into the series where n equals 1 to infinity, you're replacing x with 2, not n. And so in the numerator, remember our thing looks like 2 to the n. Sorry, it looks like x to the n over n to the n. So the denominator is left unchanged, n times 2 to the n. If we plug in 2 for x, we're going to get 2 to the n here. Notice that the 2 to the n's cancel out and we're left with the sum of 1 over n and equals 1 to infinity. This is the harmonic series and therefore this is divergent by the p-test or just the fact is the harmonic series. We know it very well. So when n equals 2, we're going to get divergent. So do not include that in the interval of convergence. But just because it didn't work for 2 doesn't mean it won't work for negative 2. It also doesn't mean it will work for negative 2. You've got to try it individually. So we get the sum n equals 1 to infinity. In this case, you're going to get negative 2 to the n over n 2 to the n. And so in this situation, you'll notice that as you cancel out the 2 to the n's, the 2 to the n cancels out but you're left with a negative 1 to the n in the numerator. So you're going to get the sum as you get negative 1 to the n over n n equals 1 to infinity right here. Now this one actually is convergent. This is the alternate or the alternate harmonic series. And so this one's convergent by the alternating series test. And so negative 2 should be included in the interval. So as our final answer, our interval of convergence would be negative 2 to 2 where negative 2 is included but positive 2 is not. And so this is our interval of convergence right here. Now for full credit on this question, we have to first identify the radius convergence. You're going to get the most of the points there. But you have to also check individually the two endpoints. So I need to see a check for positive 2 and see a check for negative 2. If you're missing those endpoints, the checking of the endpoints, if you just say that radius of convergence is 2 and stop there or you say some erroneous statement 2 comma 2, something like that. Or even if you get the right statement, if there's no visual check that you actually see why 2 diverged and negative 2 converged, then of course that's going to be forfeiture of some points. Make sure you check the endpoints on this question number 10.