Now that we've seen a bunch of theory behind Series, let's look at some of the practical questions you're going to find out how to sort of approach problems when you get them. So in this video, all I want to do are examples, and I want to walk you through how with so much information to process, you can approach at least how I approach these things. So what I want to do is talk about convergence or divergent, somewhere I do just specific series. So here's a series from N = 1 to Infinity, 5 to the negative N- 7 to the negative n. Okay, so this is a single series for the negative 9- 7 to the negative n. So can you tell me what this evaluates to if it converges at all? So this is sort of like a two-part question. Does it converge or diverge, and if so to what, okay? So when I look at this thing a couple things go through my mind so far, what do we have to work with? I have geometric series, I have telescoping series. I have the test for divergence. You gotta remember all your equations. One of the things that I'm looking at really here is 2 series. So the first thing I'm going to do when I see this, and you've seen this before in calculus math for derivatives, it's usually not, there's always some adjustment that has to get made to put in the right format. So I'm going to rewrite this thing 5 to the negative n is a little funny. I'm going to write this as 1 over 5 to the N, and I'm going to do something, so normally you put a little n under the 5, but think of it as one-fifth to the n. because remember 1 to the n is just 1. So when you see 5 to the negative n, that's the same as one-fifth to the n. And also remember Sigmas distribute over addition and subtraction. So what I'm really staring at is 2 Sigmas that there's a minus, and then a second series that I'm working with. Again, the indices don't change, so from N equals 1 to infinity of 7, 1 over 7 to n, okay? So now this helps me see the series in its proper format. So like before negative exponents, they really mean one over, and in fact are it out to 1 over 5 to the n is the same as 1 to the n, 5 to the n, which is the same as one-fifth to the n. So there's the steps that help me see. Now I have a number raised to a power. These are both examples of geometric series. Now, do you remember how to evaluate geometric series? If the number in the absolute value is less than 1, then the series converges. If the absolute value of the number is greater than or equal to 1, the series diverges. one-fifth and one-seventh are both less than 1, so these are convergent geometric series. These are convergent geometric series, so now you have to use the index to tell what the term converges to. So there's a formula in the case where it's 1 and this is one-fifth over one minus one-fifth, this is the formula when N = 1, minus one-seventh, 1- one-seventh. If it were zero, there's no numerator. The numerator is just 1 if I start at zero. Okay, so we clean all this up and I get one-fifth. 1 minus a fifth is four-fifths, minus one-seventh, 1 minus a seventh is six-sevenths, okay? And then of course we can clean this up when you have fraction divided by fraction you keep change flips, so one-fifth times five-fourths. Of course, of quarter, and you could do the same over here, one-seventh times seven-sixths. So at the end of the day the 5 cancels, the 7 cancels, and you get one-fourth minus one-sixth, and that's better known as 6- 4. You can do both time methods, so 6- 4 over 6 times 4 is 24, or two-twelfths, 2 over 24. Sorry, or one-twelfth. So final answer, this is a convergent series, really two convergent series but a single, is convergent series that evaluates to one-twelfth. Now let's look at the example of the sum from n = 2 to infinity, so it's okay. We're starting at 2, you'll see that once in a while of n over the natural log of n. Okay, new series. Haven't seen something for, let's just go through the process. So is it a number? Can I do some algebra like the other one? Can I do some algebra to write this as a number to a power? The answer is no, there's nothing you can do here. Is this going to be telescoping? So if I start writing out the terms in expanded form, am I going to see some cancellation happening? You can write out as many as you want here, there's no negative sign to cancel anything, so now it's not going to be telescoping. So then I start looking and say, okay well, what else do I have? So let's try the test for divergent. Maybe this thing diverges. You also have like, is it the harmonic series? It's not the harmonic series, so let's try the test for divergent. Okay, so what do we look at? We look at the limit as n goes to Infinity of n over the natural log of n. This is the limit of a sequence. And this is good old similar count one question. So if I plug in, if I kind of blindly plug in you get infinity over infinity. The numerator is going off to infinity and the natural log is large also goes off to infinity. So all your calc 1 bells and whistles here should be going off and you say okay This is an indeterminant form, so let's use limit rule to see what happens here. So when I take the limit again as n goes to Infinity, derivative of the numerator is 1, derivative of the denominator is natural log n is 1 / n, so 1 / 1 / n. Of course, that's better known as the limit as n goes to Infinity of just n, and this goes off to Infinity. So this limit the numerator controls. Here again, polynomials always go faster than logarithms, so the numerator controls the behavior of the function. This function gets large. Now this is bad for the limit I guess doesn't go anywhere, but this is good for us in terms of telling us the behavior of the series. Do you member what the basic test for divergence says? It says if the limit of the terms does not go to zero, then the series diverges, so therefore the series diverges, just put a little D in a box here, by the test for divergence. OK, so we have one that converges. We have one that diverges again. They're going to sort of flip on either going to do all sorts of things. Expect both to happen. Let's do another example of evaluate the following. Let's do a true false kind of question here. What have we said? The series from n = 1 to infinity of a to the n some term, I don't know this, but we're going to raise it to the 100. It must diverge. True false. Some expression a to the n 100 must divert you want you can pause the video. You think about this for a minute. So what's going on with this one? So this is for any sequence. Any term I put in here and I raise it to the 100. I know nothing about a to the n true or false? Okay, so this one is going to be false. Why? Well the problem is we don't know enough about a to the n, so the limit of the terms n to infinity of a to the n it might be 0, this could be 0. In which case, which would imply that the limit as n goes to infinity of a to the n to the 100? This would be like zero to 100. This would also be zero and therefore remember when the terms go to zero that's not enough to conclude anything, you only know by the test for divergence when the series diverges. So the problem is we just don't know anything at all. Okay, there isn't enough information about this one, so snowing that it must diverge, just not enough information. Don't know. OK, so let's try another one. So suppose that we know that the sum from n = 1 to Infinity of a to the n then we know this converges. Sometimes you see this abbreviated as the sum is less than infinity. That means whatever it is, it's not infinity. It's going to actually converge and we know that the limit as n goes to infinity of b to the n is not equal to zero. Okay, so it's not equal to zero. True or false. So given that those two things are true. Is it true that the sum from a to the n from n = 1 to infinity plus b to the n must diverge? True or false given that the sum from one to infinity b to the n converges and limit b to the n does not go to zero, then this sum must diverge. You can pause the video if you want for a minute, try to think it through. So this one is going to be true. Well it had to be true. The last one is false. Now it's going to be true just because it is true. Remember that sigmas distribute over addition like derivatives, like integrals. So we have that this really is asking about two separate series from one of them being now we know that the first sum from one to affinity and converges, so this thing converges. But we also know the limit of the bn terms don't go to 0, so we have something that converges and we're adding it to something that diverges. Now why does secondly diverge? Doesn't go zero. That's the test for divergence. So if I take some number and add it to some divergent series is like taking some number and add it to infinity. I'm going to get back divergent sum. So this thing has to diverge, so this diverges in full and we get. So the basic limit test here basically test for divergence kicks in. Okay, so there's a couple true false questions that you might get. Let's try something else here. So what is? Let's do another one where I ask you to evaluate. Let's take the sum from n = 1 to infinity of 1/n(n+1)+1/(n+2)(n+1) so big scary sum. This is all one sum and I want you to find me the sum. I'm telling you it converges, so I want you to tell me to what does it converge to. You can also. Right, now we saw one of these before the last 1 / n + 1. These look like partial fractions. So all the bells and whistles should be going off that this is prob. Probably going to be telescoping, and it's also 2 sum, even though they're asking about 1, it feels and looks like 2 so let's break this up into 2 pieces. So first and foremost, let's break off the first one and we can do partial fractions. Now I did this before so I'm going to kind of skip the work here, you can definitely check by just multiplying these together. That you can break n + 1, that you can break these off 1 over n- 1 over n + 1 you can check, that's the first one, and let's just see this one through. So if we write out the partial sums, so if you plug in 1 to get 1 minus a half and if you plug in 2, remember this one hopefully if you watched the last one, you get a half minus a third and it goes on and on forever, a third minus of fourth. So the halves all cancel the thirds all cancel, so for any finite partial sum, you always have, this will be the limit as n goes to infinity. There's always a 1 out front that doesn't cancel with anything, and the last term, last term for the first, second and third term is always- 1 and then n + 1 always. When you take the limit as n goes to infinity, the 1 over n + 1 that goes to 0, 1 over infinity, so that equals 0 and so you just get 1. So this first piece goes to 1, now let's do the second piece. It's very similar, so if you haven't done one of these before, you've only watched me do it, pause the video, and then try to work this out on your own. But it's also very similar, and again it looks like partial fractions, so that's your clue to try to play around with it and work this out. Ready, so I'll leave the details to you here, but you can definitely check by just multiplying these together. You can break this up with a and b as 1 and- 1, so you get 1 over n + 2- 1 over n plus, all the same reasons its really just picking the sequence up a little later. If you write this out expanded form, you get a third minus a fourth plus, and for the second term on there you get a fourth minus a fifth and it just goes on forever and ever. And the two inside terms, they all start canceling and very very similar to the last one. If you write out the sequence of partial sums, the first one-third stays, nothing cancels with that. And the last term which is minus 1 over m, in this case plus 3, you get this, and then when you take the limit, the one-third is a constant, so it's limited to one-third minus 1 over infinity, that's going to go to 0, so you just get good old one-third. So the first part 1 over n, n + 1 goes to 1 and the second part 1 over n + 2 times n + 3 all in the denominator goes to one-third. So put them together, when you add them up you get 1 plus a third, which is better known as four-thirds. So this series will converge to four-thirds, let's do another one, how about a true false one? So if I tell you that if the series from n is 1 to infinity of a to the n and the sum of n = 1 to infinity of b to the n both converge. So I have the sum if they both converge, then it must be true that the sum from n = one of the product of the terms am bn must converge. Get lazy at writing less than infinity, so it must converge, true or false? Pause the video if you haven't worked this out, but this one's going to be false. The reason for this is if you know that the series converges, all you know is that the limit of the terms must go to 0. So we know the limit as n goes to infinity of n goes to 0 and we know the limit of the terms as n goes to infinity of bn is also 0. Put the product together, you get that the limit of the terms an times bn, this will also go to 0, but the problem is just going to 0 isn't, the converse is not true. If I converge the limit as the terms go to 0, but if the limited terms go to 0, well then I can't conclude anything. As an example, just to give you an example, we can let an equals 1 over the square root of n. Now we don't have a test for this yet, but I promise you, so you think of it as 1 over to the one-half. If you look back at this example in a few sections, you'll see that this one is a pretty raised straight forward to see that it converges, this one converges. So the square root of n converges, this will be called the p-tests. I know we haven't done it yet, but this is just trust me on this one, but again, you can conclude anything that's the way they want you to think about at this point. But if I take an and I let this one also equal bn, if I also let it equal to bn, and so this will converge, but then their product, then you can create, what is the square root of n times square root of n? Well, that's just n, so the product of the term Is 1/N, and of course 1/n, that series, we know this is our famous harmonic series and this guy diverges, so it's possible to cook one up where that happens, okay? All right, so tough one here, but this one is four, as another one, here we go. So let's look at, if the series, if the limit, as n goes to infinity of a to the n, b to the n is 0, then which of the following are true? Let's write it this, also give us a multiple choice question a, that the sum from 1 to infinity of a to the n converges. b, that the sum from one to infinity of a to the n, b to the n must converge. c, that the sum from n= 1 to infinity of b to the n converges, or let's say d, none of the above. Take a second, work this out if you can. All I know is that the product, an, bn, their limit goes to 0. Okay, which of the following are true? Can I conclude that the first term goes to 0, is that enough? Just knowing that the product goes to 0. By the way, this means that an might be constant and bn might be going to 0 or infinity, 1 over infinity go to 0, who knows. But the point is, if the limit goes to 0, again, you can't conclude convergence, if you're trying to apply the test for divergence. So that's not true, can't conclude that. For all the same reasons, just because the product goes to 0, I can't conclude that the series converges, and for the same reasons why a is false, c is false as well. So this is one of those rare examples where the answer is in fact none of the above. Let's do another one, this will be straight up evaluation one. So I want you to evaluate the series from 1 to infinity of 1 + 5 to the n, plus 3 to the n, all over 7 to the n. Pause this one, take a second and work it out if you can. But when you look at this, of course, it's in a weird format, so of course we're going to rewrite this. So let's break this up into a bunch of pieces. This is really three series in disguise here, so we'll break up the fraction. We've seen this before, so 1/7 to the n, plus the sum from n= 1, so 5 to over 7 to the n, and plus, last but not least, n= 1 to infinity of 3/7, so this is really three series. They're all convergent geometric series. We have a number to a power, and because the first index, any series is 1, we can use the formula. Remember there's a formula that says if I have the index is 1, so we usually write it as r to the n, when n =1, and assuming that it converges. This is r/1- r, when the term is 1, that's important, that's what we have now. So let's write this out. So this becomes 1/7, 1- 1/7, plus 5/7, 1- 5/7, plus 3/7, 1- 3/7. Now, we just work it all out, so 1/7 times, let's do 1- 1/7, that's 6, 7, so 7/6. Yeah, 5/7 and then times, what would that be? 7, 7, 2, 7, 7/2 plus 3/7, and then you have, 7- 3 is 4, so 7/4, all right. So lots of canceling going on, with those sevens, and you get 1/6 + 5/2 + 3/4, and I leave this to you to workout the arithmetic. But put them over common denominator of 12, and you get 1, 2, 30 and 9, so it's 41/12. But you can check, okay? So that converges, so nice. The crazy but good example of a geometric, let's do one more. Let's try to make this as weird as we can get it to. So you're not surprised if you see these things. So if I tell you that the sum from k is 1 to infinity of k 2/3 to the ks, and now my index is k, okay? Who cares? Whatever, I tell you that this is equal to 6. Then What is the sum from k is 1 to infinity of 10k- 3, time 2/3k, minus the sum of k is 1 to infinity, 5K minus 3 2/3k, times 2/3k. What does that equal to? If you could do this one, you could do any of them. So remember, pause the video, try to work this out. To do this, you look at this crazy double. Come over here and you gotta remember to yourself that wait a minute, sigmas distribute over addition. So in particular these two sigmas not only distribute, but they also factor, so you can factor out, so what do they have in common. They have a two thirds k in common so let's pull that out front, two thirds k. And then what I'm left with is 10k minus three and then minus 5k minus 3. So this is the first observation to make, and k is 1 to Infinity. And then, let's see what happens now. So if I ever keep going, I have to write equal sign every time and write sigma every time, I get 10k minus 5k. Well that's not too bad. Then I got 3 minus 3 plus 3, okay so this just 5k. When you clean this up you get two thirds raised to the k and then times a 5k. Five is a constant. You can bring it out front, so you can almost think of this as five times the sum from k is 1 to infinity. Of 2/3 k Times k. I say wait a minute. That looks familiar because we have this already right? Look at that, from K is 1 to infinity, 2/3 k is k is 1. It's all here. It's given, this piece of information is given. They tell us that this is equal to six. So this becomes 5 times 6 which is equal to 30. This question is tricky because most, I guess you could go that route, but people try to find the value of k first, but it turns out you don't actually need it. You don't actually need it, so whenever you see something horrible and gross, I know it's going to look scary at first, but almost get excited about it has to, it's going to clean up. Something's going on, so try to look for a shortcut if you can. All right. Good job on these. Try some more. You can't do enough of these problems. Get the geometric series. These are the ones that are going to show up a lot, geometric. Telescoping a little less so and harmonic a little less so. But know them when you see them. Especially because they're so nice. They're our friendly series. They're our ones that converge, and we have formulas when they converge. So we can tell you what they converge to. Not going to happen always as we'll see in the next video. So work on these and I'll see you there.
