 We have seen previously that given a power series There is one of three possibilities that happens with its interval convergence The radius of convergence turn could turn out to be zero for which the series only converges at its center The radius of convergence could be infinity for which this that means the interval convergence will include all real numbers or We could have that our radius of convergence is a finite number. That's positive So we have some finite interval of numbers for which the series converges on So I want to work through an example of such a thing and let's identify the radius and interval of convergence So you always want to start off with the radius of convergence and you're going to want to use the ratio test To help you out here. The root test could also be useful But in practice, I'd say the ratio test is going to be typically more efficient So we have to look at the sequence that we're adding up in this series. So we're looking at a n plus 1 over a n like so in which case We have to compute the limit of ratios and determine when does this thing less than 1 well for this series if we look At a n plus 1 we're going to get negative 3 n plus 1 times x the n plus 1 Over the square root of n plus 2 Notice n plus 1 plus 1 is n plus 2 and then if we take the reciprocal of a n We're going to get the square root of n plus 1 Over negative 3 to the n times x to the n and let's combine some terms here We have a negative 3 to the n plus 1 we have a negative 3 to the n on the bottom This will cancel out with most of the negative 3's on the top for which case if we take the absolute value of negative 3 We're going to get a positive 3 so we're going to get 3 right there That's pretty nice The next thing is to look at the powers of x we have an x to the n on the bottom That'll cancel with most of the x to the n plus 1's on the top There'll be one x left over so we get an absolute value of x for which we'll record right here And then the last thing to do is put together your square roots You're gonna have the square root of n plus 1 over n plus 2 Notice that this square root is the only part that depends on n and so as n goes to infinity right here So we're taking the limit as n goes to infinity We end up with 3 times the absolute value of x times the square root of 1 n plus 1 over n plus 1 over n plus 1 is a balanced rational function That'll go towards 1 and as the square root function is continuous We'll get the square root of 1 which is itself 1 so this becomes 3 times the absolute value of x And then we have to solve the inequality. When does this thing? When is this expression on the left less than 1 because that is when the ratio test guarantees convergence Solving for this we divide both sides by 3 We end up with the absolute value of x is less than one-third which tells us that x will be less than one-third But greater than negative one-thirds we get this we get that right there So now going back to the original question about the radius I want to mention that this statement right here is what we were looking for Once we get that the absolute value of x is less than a positive number or the absolute value of x minus a If the center is something other than zero if the absolute value of x minus a is less than something that number is going to be the radius of convergence So we can see right here that our radius of convergence here are is going to equal one-third So we found the first bit that with our center being zero We can go one-third to the right one-third to the left and there's guaranteed convergence there So our radius of convergence is going to be one-third We know our series will converge when x is greater than one-third It'll diverge when the x is less than one-third the absolute value of x is one-third of course So but what about if the absolute value of x equals one-third exactly what happens in that situation? So this we have to be a little bit more careful about when x equals one-third We have to consider the sum when n equals zero to infinity We're going to get negative three to the end, but then we're also going to get a one-third to the end over the square root of n plus one and notice how There's going to be some combination going on right here This thing when you take three to the n times one-third to the end That's just going to give you one to the end aka one and there is this negative sign that sticks around So what you can see it happening is you actually get a sum as n equals zero to infinity You're going to get a negative one to the end over the square root of n plus one right here and so this series is an alternating series and In fact by the alternating series test This series is going to be convergent Notice that the sequence one over the square root of n plus one Decreases toward zero and so the alternating series test gives us convergence So in terms of our interval of convergence We know that x so for our interval of convergence x sits between negative one-third and one-third One-third is included. What about negative one-third? Well, the only way to know it for sure is we have to plug in x equals one-third into Thus power series and see what happens I don't want you to assume that just because one side converts the other side necessarily diverges It turns out like we saw on this slide right here Both sides could could could converge It could be that the right side converges and not the left side the left side converges But not the right side or it could be that neither the left or right converges We don't know without investigating it individually. So let's try it this time if we plug in x equals negative one-third Into our power series we get negative three to the end times negative one-third to the end over the square root of n plus one Like so n equals zero to infinity Well, this right here when you simplify the numerator actually will just become the sum of one to the end Over n plus one square of n plus one. So we're just getting the sum of one over the square root of n plus one And so when I look at something like this This tells me that most likely this series is divergent because this kind of looks like it's a This kind of looks like it is a p-series right one over into the one-half power You have to do some type of comparison test right removing the plus one on the bottom makes the denominator get smaller a Smaller denominator makes a bigger ratio And therefore we actually could show that this series I have to make sure I said this right so if you remove the plus one the denominator gets smaller Which means the fraction gets bigger right and so since the fraction gets bigger We could we could actually make an inequality symbol right here. This would be oh No, I think I keep on saying it backwards there. I'm sorry everyone if you give it of the plus one for right here that makes Well now now I'm just embarrassing myself, right if you get rid of if you get rid of the plus one on the bottom Small that that's a smaller denominator makes a bigger fraction right so that actually I'm sorry that means this one's bigger So that that's why I'm getting a little bit confused here. I what I was saying was was correct I wanted to point the inequality in the wrong direction So what we have here is that this p-series is actually bigger So we can't make it we can't make a straightforward comparison test But you could easily do a limit comparison test and so since this is a p-series Which is divergent now it implied that this one is divergence as well And so since x equals negative one-third diverges This wouldn't tell us that we're going to get a Prothesis right here. And so our interval of convergence will be negative One-third to one-third where one-third is included but negative one-third is not included so we can compute both the radius of convergence and the interval convergence and Most of your calculations for radii and interval of convergence will look very similar to what we see in this example here