Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. An infinite sequence of numbers is a function whose domain is the set of positive integers. The sequence a sub N converges to the number L if to every positive number epsilon there corresponds in integer N such that for all N, N is greater than that integer Implies that the absolute value of your a sub NTH term minus the limit, is less than the epsilon or the error value. If no such number L exists, we say that a sub N diverges. So basically what we're looking at is we're looking at a single term and we're trying to figure out if that term, when N is large enough, goes to a number or goes to Infinity. If it goes to a number, it converges, if it goes to Infinity, it diverges, or it could oscillate and still diverge. So if a sub N converges to L, we write it as the limit as N goes to Infinity of a sub n = L, or simply a sub N implies L call L the limit of the sequence. Remember, we're looking at a single term. Eventually we will look at a series where we're actually adding up all of those individual terms to find out what happens to the summation. So the sequence square root of N diverges. Thus the limit is N goes to Infinity of the square root of N equals Infinity. If N is really really large, the square root of something really really large is still really really large. The sequence a sub N diverges to Infinity if for every number M there is an integer N such that for all N larger than that integer, a sub N is greater than some M. Thus the limit is N goes to Infinity of a sub N equal Infinity or a sub N implies Infinity. If we let a sub N&B sub N be sequences of real numbers, and we let A&B be real numbers, the following rules hold. If the limit is N goes to Infinity of a sub N equals capital A, and the limit is N goes to Infinity of B sub N equals capital B, then the sum rule says the limit is N goes to Infinity of those 2A sub n + b sub N is going to equal the integers A+B, actually the real numbers A+B. The difference rule holds also the quotient rule once again, although this time we have to make sure the denominator doesn't equal 0. The product rule, the constant multiple rule and the constant multiple rule just basically says if I have a constant inside a limit, the constant can come in and out. We have the sandwich theorem here for sequences. So if a sub N&B sub N&C sub N be sequences of real numbers, if a sub N is less than or equal to B sub N which is less than or equal to C sub N holds for all N beyond beyond some index N. And if the limit is N goes to Infinity of a sub N equals the limit is N goes to Infinity of C sub N equal L, then we know that whatever that term in the middle was has to also go to the L as N goes to Infinity. We can also say if the absolute value of B sub N is less than or equal to C sub N&C sub N goes to 0, then B sub N has to go to 0. If it's the absolute value of B sub N, we know it's got to be in between negative C sub N and positive C sub N. And if C sub N goes to 0, then the opposite of 0 is still 0 and hence by the sandwich theorem or the squeeze theorem that B sub N has to also go to 0. A few key convergent divergent sequence limit as N goes to Infinity of 1 / n one over something really really big goes to something really really small, IEA number, and hence it converges. The limit as N goes to Infinity of some K. There is no N in here, so no matter what our N is, we're always going to get out K, so that one's going to converge. Also, K being some constant, the limit as N goes to Infinity of -1 to the n + 1. This is going to diverge because it's going to oscillate. It's going to go one negative 1/1 negative 1/1 negative 1. So it has to diverge because it's not going to a single number cosine n / n. If we think about the fact that cosine N had to be in between -1 and one, and then if we divide each of those 3 by N, we know that the limit as N goes to Infinity of -1 / n is 0, and the limit as N goes to Infinity of 1 / n is also zero. So by the sandwich theorem that this one in between has to also go to 0. Thus it converges. If we look at the limit as N goes to Infinity of 1 / 2 to the N power, we could think of this as 1/2 to the N power because one to any power is going to be one. Well, 1/2 to the N. We could think of that being greater than 0, but it's less than 1 / n. And since we know that the limit is N goes to Infinity of 1 / n goes to 0, we know that by the sandwich theorem, this 1/2 to the N power also has to go to 0. If we look at -1 to the n * 1 / n. Once again, doing the sandwich theorem, if we look at -1 / n and 1 / n because we know that -1 is going to be less than or equal to -1 to the N which is less than or equal to 1. So if you start with that idea, think about starting with this, we know that that's going to be true because this -1 to the N is always either -1 or one. And then we're just taking each term and dividing it through by an N And then we showed earlier that the two end points as N goes to Infinity are both going to 0. So by the sandwich theorem, the middle one has to go to 0. We have limit as N goes to Infinity of sqrt n + 1 / n If we rewrote this, we'd have 1 + 1 / n. And as N goes to Infinity, 1 / n goes to 0, so we get equals 1 and thus it converges. Remember, these are just some general standard ones that we're going to use to build upon to develop more difficult ones. The limit is N goes to Infinity of the natural log of n / n. This time we're going to use L'opital's rule because it's an indeterminate form. So the derivative of LNLNN is 1 / n and the derivative of N is one. One over something really, really big goes to 0, thus it converges. The limit is N goes to Infinity of two to the north over 5 N Once again, we're going to use Lopital's rule. We're going to get 2 to the NLN 2 / 5. Two to something really, really big is going to go to Infinity. So we're going to diverge the NTH root of N This one's a little more challenging to see. We could think of the NTH root of N as the limit is N goes to Infinity of N to the 1 / n But this really is an indeterminate form because if we have think about Infinity to the 0 power, we don't know what Infinity to the 0 power is. What we're going to do is we're going to let M equal that N to the 1 / n. Taking the natural log of each side and bringing down the exponent, we'd get the natural log of M equaling the natural log of n / n. Now if we think about what happens to this as the limit of N goes to Infinity, we'd use Lopita's rule and we'd get 1 / n / 1 limit. As N goes to Infinity would give us zero, so the natural log of M is going to equal 0. Changing this back into exponential form so we can get our M by itself, we get M equaling E to the zero or M is just one. So the limit as N goes to Infinity of the NTH root of N is equal to the limit as N goes to Infinity of one. And by the constant rule, that's just one, and thus this one does converge. The limit is N goes to Infinity of X to the 1 / n equal 1 when X is greater than 0. So one over something really, really big is going to get really, really small. If X is a positive number greater than 0 and we're taking it to something really, really small, we going to get out. One limit is N goes to Infinity of X to the N power equaling 0, the absolute value of X less than one. This time we have X being in between -1 and one basically, or a fraction. So if I have a fraction to something really really big, that's going to give me out zero because a fraction to something really really big, we could think about taking the reciprocal of it and that would make it to a negative exponent and hence it goes to 0. It converges. The limit is N goes to Infinity of 1 + X / n to the NTH. This is a definition. This is what E to the X is defined as. So it does converge for any X. The limit is N goes to Infinity of X to the end of the N factorial is going to equal 0 for any X. Thus, it's also going to converge this one visually. If we think about X to the north, we have X * X * X * X times XN quantities of time. And then we have N factorial. So NN -1 N -2 N -3 dot dot .321. These XS are all constants, but this N is really, really big. So a constant over something really, really big times a constant over something pretty big times a constant over something pretty big. Eventually we're going to have a constant over one, which is still just a constant, and a constant over 2, which is still just a constant. So we're going to get something really, really big in this denominator, and hence it's going to go to 0. Now, if we actually look at some examples, a sub N equal 1 / n factorial. We want to list out the 1st 4 terms, so we're going to put in a sub one being 1 / 1 factorial. By definition, 0 factorial is one, and one factorial is also one. Those are definitions that if you don't know, you should become familiar with. So a sub two is 1 / 2 factorial or 1 / 2 * 1. A sub three 1 / 3, factorial 1 / 3 * 2 * 1. Ace of four is 1 / 4 factorial 1 / 4 * 3 * 2 * 1. Perhaps we should talk about N factorial means n * n - 1 * n - 2... Times 3 * 2 * 1. This next one, they actually give us a listing of numbers and want us to come up with the a sub N So the first thing is it's oscillating between positive and negative, positive, negative. So we're going to have -1 to the north plus one, and we can see that we could think of one as 1 / 1 and 1 / 4, one over nine, 1 / 16, one over 25. So that's going to be 1 / n ^2. Now there are lots of different possibilities, just just happens to be one of the easier ones to see. We could have had a sub N equaling -1 to the n - 1 / 1 / n ^2, or lots of other possibilities, but that's one of them. The next example is called a recursive definition, and what recursive definition states is that the next term is in terms of the previous term. So in this example, we're given a sub one is 2A sub 2 equal -1, and a sub n + 2 is a sub n + 1 / a sub N. So when I want to find a sub three, that would be my next term. If I want a sub three to get three, we're going to have the N being one because 1 + 2 is the three. So everywhere I see an N, I'm going to put in the #1A to the 1 + 1 / a sub one, or a sub 2 / a sub 1A sub two is -1. A sub one is 2, so a sub 3 is negative 1/2. To find a sub four, this 4 has got to be n + 2. So this time the N is going to be two. Every time I see an N, I'm going to put in the two now. So a sub 2 + 1 / a sub two, or a sub 3 / a sub two negative 1/2 / -1 or a half. A sub 5 = a sub 3 + 2. So a sub 3 + 1 / a sub three or a sub 4 / a sub three. In this case, it's going to be a -1. If we do a sub six, the N is now 4. When we do this, we get -2. A sub 7 is just a sub 6 / a sub five, which would be a positive 2A sub 8 is just a sub 7 / a sub 6A sub 9 is a sub 8 / a sub seven. We could keep going on and on and on. The next one, a sub N equal n + -1 to the N all over N. We want to decide whether this is going to converge or diverge, so we're going to look at what happens to the limit as N goes to Infinity of the single term. We know that n / n is 1 and -1 / n negative 1 to the n / n so we're going to split this up using that sum rule. So the limit is N goes to Infinity of one that's a constant, so it's just going to be 1. And the limit is N goes to Infinity and -1 / -1 to the n / n we showed a minute ago is really just zero by doing the sandwich theorem. Remember we had -1 less than or equal to -1 to the N which was less than or equal to 1. And then we divided each of those by N and we know that the two NS are going to go to 0 if N goes large enough, so the middle one has to also go to 0. So that term is going to converge A to the north of -1 to the north times 1 -, 1 / n this -1 / n is going to oscillate back and forth between negative and positive. And this piece here is going to go to 1. So we're going to get -1 and then one and -1 and one. So it diverges because the limit does not exist, because when we put in N going to Infinity, it's going to oscillate between -1 one, negative 11, etcetera. If we have a sub N of sine squared n / 2 to the N first thing, sine squared N has got to be in between zero and one because plain old sine has got to be in between -1 and one. So if I square the sine squared, now we're going to be between zero and one. If I take each term and divide by two to the N2 to the N2 to the N, as N gets really really big, this is obviously going to 0. If N is getting really really big, this is also going to go to 0 by the sandwich theorem. Then the middle has to converge to 0. Thank you and have a wonderful day.