Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. The ratio test Let the summation of a sub N be a series with positive terms, and suppose that the limit is N goes to Infinity of a sub n + 1 / a sub N equals some P. Then the series will converge if the P is less than one. The series will diverge if P is greater than one or if P is infinite and the test is inconclusive if P = 1. We're going to use the ratio test frequently when we have factorials or expressions raised to the N power. The root test. We're going to let the Series A sub N be a series with all the terms greater than or equal to 0 for N greater than or equal to some N. Hence the beginning terms might be negative, but when we get out far enough, they all have to turn positive. And suppose that the limit is N approaches Infinity of the NTH root of a sub N equals some P. Then the series converges if P is less than one, and the series diverges if P is greater than one. It's inconclusive if P = 1. We're going to look for some patterns. So recall that limit is N goes to Infinity of X to the 1 / n = 1. When X is greater than 0, limit is N goes to Infinity of X to the n = 0. When the absolute value of X is less than one, the limit is N goes to Infinity of the quantity 1 + X / n to the N power is E to the X and the limit is N goes to Infinity of the NTH root of n = 1. So if we look at some examples, sometimes we can do problems in multiple tests. So if we look at N equal 1 to Infinity of n ^2 * e to the negative N, we're going to do the ratio test. So the limit is N goes to Infinity of a the n + 1 / a sub N So we'd have n + 1 ^2 e to the negative n + 1 / n ^2 e to the negative N. So if we have two things that are squared, we could write it as a fraction as the quantity squared. And if I have E to the negative n + 1 / e to the negative N, those bases are the same. So we can subtract the exponents and we'd get 1 / e or E to the -1, which is 1 / e. Now if I take that inside parenthesis and split it up n + 1 / n into 1 + 1 / n ^2, that's really just going to go to one as N goes to Infinity because the 1 / n is going to get me zero and 1 + 0 ^2. So then we're going to multiply it by 1 / n, and that's going to equal 1 over. Sorry, we're going to multiply it by 1 / e. That's going to equal 1 / e, which is going to be less than one. Thus it converges. Now, another test that we could have used is we could have done this exact same problem with the root test. If we did the root test, we'd have the NTH root of n ^2 e to the negative N We know that the NTH root of N is really just one, so 1 ^2. And then we'd have the NTH root of E to the negative N. Well, the NTH root of E to the negative N would really just give us E to the -1 1 ^2 * e to the -1 is 1 / e, which is less than one again, and hence it converges. Just to refresh and remember that the 8th A root of X to the b = X / b to the A. We could also think of doing it as the 8th root of X first, all raised to the B power. So those three things are equivalents. Looking at some more examples, if we have N equal 1 to Infinity of N factorial over 10 to the N, whenever we see a factorial at this point, we're going to want to try the ratio test. It won't always tell us anything, but that's going to be our test of choice for factorials. So we have n + 1 factorial over 10 to the n + 1 / n factorial, 10 to the N. So if we take the reciprocal of the denominator and multiply, we get n + 1 factorial over 10 to the n + 1 * 10 to the n / n factorial. Remember that 10 to the n + 1 we could really think of as 10 to the n * 10. So those 10 ends on top and bottom are going to cancel. Now if we look at this n + 1 factorial, n + 1 factorial is really n + 1 times n * n - 1 * n - 2 all the way down to 3 * 2 * 1. This N factorial is n * n - 1 * n - 2 all the way down to 3 * 2 * 1. So all of these terms here are going to cancel with all of these terms up here leaving just an n + 1 in the numerator. So this whole thing is really just going to reduce to n + 1 / 10, and when N goes to Infinity, we'd get Infinity over 10 or Infinity, which tells us by the ratio test this is going to diverge. If we look at another example, they can get very complex N equal 1 to Infinity of N2 to the NN plus one factorial over three NN factorial. If we do the ratio test again, we would see that we have n + 1 two to the n + 1 N +2 factorial because n + 1 + 1 / 3 to the n + 1 N plus one factorial. I'm going to go ahead and just flip it and write in this next term. So the three to the NN factorial over N2NN plus one factorial. So all sorts of things are going to happen here. When we look at 2:00 to the n + 1 / 2 to the N, we subtract the exponents and we'd get a plain old 2. If we look at 3:00 to the n / 3 to the n + 1, we'd subtract the exponents and we'd get a three at the bottom. If we look at n + 2 factorial and n + 1 factorial, we could think of n + 2 factorial as n + 2 * n + 1 factorial. So the n + 1 factorial on top would cancel with the n + 1 factorial on bottom, leaving us just to n + 2. If we did the same thing, looking at this n + 1 factorial and the N factorial, this n + 1 factorial is going to be n + 1 times N factorial. So the N factorial portions cancel, leaving me an n + 1 in the bottom. And then I still had this n + 1 on top and an N in the bottom. So the N plus ones on top and bottom will cancel. So this actually reduces to 2 * n + 2 / 3 * n. Now we could do Lopital's rule here because Infinity over Infinity it's in an indeterminate form. Or we could think about multiplying everything by 1 / n It doesn't matter if we just do Lo Patel's rule where we get 2/3. So the limit is then it goes to Infinity of 2/3 or of a constant is just 2/3. So it converged just by the ratio test. I could have also shown you if we divided everything through by N, we would have had two times the quantity 1 + 2 / n all over three. It's the same thing as doing Lopital's rule basically. So 2 * 1 + 0 / 3 is 2/3. So this converges by the ratio. On this next example, N equal 1 to Infinity of 1 / n - 1 / n ^2 to the NTH power. Whenever we have an NTH power, we want to use the root test if we can, because the NTH root of the NTH power are going to cancel. So we look at limit as N goes to Infinity of 1 / n - 1 / n ^2. Now these are terms. Remember, these aren't series, so we actually have to evaluate what this limit as N goes to Infinity is really doing. Getting a common denominator, we can see n -, 1 / n ^2. If we take Lopital's rule, we'd get 1 / 2 to the N which would equal 0. And by the root test when it equals zero, we know it converges if we look at N equal to to Infinity of N over the natural log of N to the n / 2. Once again, taking the NTH root or the root test, the NTH root of N is just one, the NTH root of L&N to the n / 2. The ends will cancel, leaving us the natural log of N to the 1/2. That being in the denominator. When we put in an Infinity, we're going to get something really, really big in the denominator and it's going to go. The whole term is going to then go to 0, thus it converges. If we have a recursive equation, we want to always use the ratio test if possible. So if I have a sub one equal 5 and a sub n + 1 equal the NTH root of n / 2 a sub N, If I use the ratio test it says I want the a sub n + 1 over the a sub N and what will happen is those a sub NS will then cancel. So now we get the limit as N goes to Infinity of the NTH root of N which is 1 / 2. The fact that it's 1/2 means that it's going to converge. If we look at this next one with the ratio test, we can see a 1 equal 3A sub n + 1 equal n / n + 1 a sub N Using the ratio test, there's a sub NS cancel, we get n / n + 1. Lopitas rule we get 1 / 1 which is one. This is inconclusive, so we cannot use the ratio test. So the next thought process might be to list out some of our terms. If I get a sub one is 3A sub 2 was a sub 1 + 1, or 1 / 1 + 1 * 3 or three halves. A sub 3 is really a sub 2 + 1. So every time we see the N of two, we're going to or N we're going to stick in two. So we get 2 / 2 + 1 * 3 / 2 in this case because the term right before it, that would equal 1A sub four. Every time we see the end, we're going to stick in three. So 3 / 3 + 1 times the a sub three term, which was 1. So that's going to give us 3/4. So if we have three and three halves and 1 3/4, we might think of that one as 3 / 3 so we could get the threes all consistent in the numerator. Then we might see the pattern of 1234 plus dot A dot. SO N equal 1 to Infinity of 3 / n. Well we can pull the three out in front. So we get three times the summation N equal 1 to Infinity of 1 / n. While this 1 / n is really our harmonic series and we know that the harmonic series diverges. So a constant times the harmonic series is still going to diverge. So the original function diverges. If I have a sub one equal 1 and a sub n + 1 equal 1 plus tangent inverse n / n * a sub N. Doing the ratio test we have the a sub n + 1 / a sub N. So the a sub NS are going to cancel and we get the limit as N goes to Infinity of one plus tangent inverse of N all over N Remember our tangent graph? If we think about what happens when the Y goes to Infinity, we would say that the X is going to Pi halves. So this tangent inverse N is really going to Pi halves. So we get 1 + π halves all over N. Well, as N goes to Infinity, this whole thing then is going to go to 0. So if it goes to 0, that's less than one. Thus it converges by the ratio test. If we have N equal 1 to Infinity of N to the NTH over 2 N squared root test 2 to the n ^2. We're going to use the root test because we have these NTHs in the powers. So the NTH root and the N powers are going to cancel. So we get limit as N goes to Infinity of N and if we have the NTH root of 2 to the n ^2, we could write that as n ^2 / n or just plain old N. Now if we do L'opital's rule, we'd get 1 / 2 and natural log of two as NS getting really really big. We get one over something really really big, which goes to 0. Thus it converges by the root test. Thank you and have a wonderful day.