 Hello, this is a video about using the ratio test to test for convergence of infinite series So first what is the ratio test exactly? Well, we're gonna let a sub n be a series with nonzero terms The first component of the ratio test is that the series converges Absolutely when the limit as n goes to infinity of the absolute value of the ratio of a sub n plus 1 Over a sub n is all less than one We say that the series diverges when the limit as n goes to infinity of the exact same ratio With an absolute value is greater than one or has a value of infinity The ratio test is in fact inconclusive when the limit value is equal to one Before we dive into using the ratio test, I want to talk to you a little bit about factorials So by definition if you see something like four factorial, it's not like saying four Like you're excited. It's actually a mathematical operation It means you take four times three times two times one It's a decreasing product by one number less each time until you get all the way down to one Which would mean five factorial would be five times four times three times two times one And so one way we can kind of write this a little bit differently is to just Write the five separately by itself and notice what the sequence product here the four times three times two times one is That's actually four factorial So we can write it five factorial as five Times four factorial we can break off that first factor and write it separately So in general this means in factorials in times in minus one times in minus two all the way until you get down to a value of one So which means that in plus one factorial would be in plus one Take away one so times in take away one times in minus one take away another one in minus two all the way until you Get down to one Equivalently we can write this as The first factor in plus one Then group everything else together. That's actually what in factorial is So we need this trick of sometimes writing out the first term or the first factor. I should say Whenever we're using the ratio test So let's use the ratio test to determine whether the following series converges or diverges You have the series from n equals one to infinity of n to the n over in factorial notice Your ace of n is the exact formula n to the n over in factorial while a sub n plus one is n plus one to the n plus one power Over n plus one factorial. I just replaced everywhere. There was an I replaced it with n plus one So first let's calculate the absolute value of the ratio of ace of n plus one over ace of n So what happens here is I have my ace of n plus one written first And then I'm dividing by ace of n remember when you're dividing by a fraction. You're really multiplying by the reciprocal So that's what I did here. I'm multiplying by the reciprocal of ace of n in factorial over into the n Now the goal here is always to try and get some stuff to cancel out So what I did first was I took my n plus one to the n plus one power I Broke it up into n plus one to the n and then n plus one to the first it's kind of like using the Properties of exponents in reverse think about what would happen if you were to recombine the n plus ones here You're multiplying your bases are the same you add in and you add one to get in plus one Similarly, I'm going to write the n plus one factorial as I did on the previous slide n plus one factorial will become n plus one write the first factor separately then in factorial Then you have your n factorial over into the n at this point. I can finally cancel out the n factorials cancel out The n plus one on top with the n plus one on the bottom of the first fraction also cancels leaving me with absolute value of n plus one to the n power over into the n Notice that for n equals one to infinity you always have a positive quantity And also notice that since both the top and bottom are raised to the n power I can rewrite as n plus one over n to the n power I will now take the limit as n goes to infinity of this ratio giving me limit as n goes to infinity of n plus one over n to the n power So I'm going to take this quantity and break it up over the denominator to get n over n n plus one over n all to the n power so This gives me one plus one over n to the n power which From l'hôpital's rule or using l'hôpital's rule. This has a value of e which is greater than one. Remember e is 2.7 18 It's greater than one Therefore we can conclude by the ratio test That the series diverges All right in our next example. We have the series n equals one To infinity of negative one to the n power square root of n over n plus one So this is actually an alternating series So notice that your a sub n is the exact series formula that they give you And then a sub n plus one is negative one to the n plus one square root of n plus one over n plus two So first we'll calculate the ratio with an absolute value. All right, so what this means for us is We're taking the absolute value of negative ones raised to powers Well, when you take the absolute value of quantities with negative ones in them Absolute value kills the negativity. So I'm going to exclude those negative one to the powers As I write this ratio out. There's no need to throw it in there It's extra absolute value kills those negative ones raised to powers. So I have square root of n plus one over n plus two I'm dividing by a sub n. So I'm multiplying by its reciprocal So this gives me n plus one over square root of n And there's not much you can do to simplify this but you do get For n equals one to infinity you always have a positive value for every single grouping So there's no need to include absolute value So I'm going to write the square root groupings all in the same fraction together and then the non-square root groupings in another fraction together And this gives me square root of n plus one over n and then n plus one over n plus two power We now have to take the limit as in goes to infinity of this quantity All right, so I'm going to take the limit As in goes to infinity of This ratio that I found you have square root of n plus one over n And then you have your n plus one over two So remember When you're taking infinite limits I want to go under the square root pick the highest power term from the top the highest power from the bottom That gives me square root of n over n Second grouping highest power term from the top is in highest power term on the bottom is also in giving me n over n This can be nicely simplified the square root of one times one which has a value of one So limit as in goes to infinity of one is indeed one This means the ratio test is actually inconclusive a value of one means the ratio test is inconclusive We can't say anything about the convergence or divergence of this series currently As a matter of fact, this is actually alternating series the alternating series test would tell you If the series converges or diverges, so thanks for watching