 Welcome back. We are lecture 38. If I remember seeing that right. Okay, good. That was a good guess Today we are going to begin section 8 to which is on series So the same kind of stuff we were looking at yesterday, but we're just looking at individual terms We're now going to be looking at the sums of those terms in that patterned array. Whatever that array happens to be I Don't know that we'll finish with this section today. Probably not But we'll just see how it goes. I think some of the problems, especially the first time through they do kind of Go slowly because we're developing formula as we Attack different kinds of problems once we develop those Algorithms those shortcuts. I think problems from that point go quicker So we are going on to series Series means that we'll have the same kind of animals except we'll have a plus sign in the middle so we've got a first term added to a second term and They may be as such that we stop out here at a sub n or they may in fact Continue so we could have infinite series infinite geometric series. In fact, that's one of the ones that we'll cover today So we're going to be looking at Things and a lot of times these will be given to us in a sigma notation kind of a closed Notation as opposed to an expanded notation where we have We don't really have a description of what this is but let's say a sub n and in goes from one to eight or If we want to let it go on We'll lay that eight on its side and make it an infinity So there is a way to figure out the sum of some series that go on indefinitely They have to be certain types in order for us to say that they converge And these that have a specific starting point and a specific stopping point Certainly, they will always converge because you're always going to be able to add them together First example, let's look at And we're not going to put this into a category as much as just kind of Examine how to tie sequences together with series Let's start this thing at one and let's let it go one over two to the end and Starts at one and we're going to let it go to infinity one of the decisions we want to make is is it convergent and If it converges and we're able to do so can we find the number to which it converges? So let's write out the kind of what this looks like the first term is what? one half second term is one fourth One eight that look right and we're going to let this thing go forever It is geometric. Why is it geometric? So by the end of class today? I think if we get this far you'll have a very good handle on recognizing when a series is an infinite geometric series and Deciding if it converges and also the value to which it converges So that'll be one of the the goals today. Why is it geometric? What has to happen in a geometric? progression or in this case a geometric series We multiply by something as we progress so as we go from here to here From here to here and so on we've got the same number used as what they call a common ratio What is the r value in this problem? It's one half. So you start with a first term and we'll kind of Categorize this a little better, but then once whatever the first term is First term doesn't have to be the ratio which it is in this case But in getting from the first term to the second term you multiply by r in Getting from the second term to the third term you multiply by r and so on if that's the case It's a geometric series. This is an infinite geometric series because it continues indefinitely Series obviously because we're adding these things together So how do we tie together? Sequences from 8.1 to series in 8.2 and really the rest of this chapter There is a sequence created if we take what are called partial sums Called pretty appropriately the sequence of partial sums So we look at the sum and I know this is not a very exciting exercise the sum of the first one terms of this series That's a toughy The sum of the first one terms would be one half Then we go on the sum of the first two terms Would in this case be three-fourths The sum of the first three terms would be what? Seven-eighths and we can continue that and we generate this sequence of partial sums We're not going to add these sums together because they are already sums in and of themselves So we generate the sequence the first term of the sequence is one half The first or the second term of the sequence is three-fourths The third term is seven-eighths and then we analyze the sequence of Partial sums to see if the sequence converges if the sequence of partial sums converges Then this series converges What would the next term of the sequence be? What's the sum of the first four terms? Fifteen-sixteenths is that right? This is our first example so we can just kind of make some guesses What do you think do you think this sequence of partial sums appears to be converging on a Certain number and if so, what does that number appear to be? It appears to be getting closer and closer to one and we can continue the sum of the first five so we'd be adding in one over 32 What's the sum of the first five? 31 over 32 and you kind of see the pattern now that we're kind of doubling the denominator and the numerator is just one one behind that denominator so the next term would be 63 over 64 so we do have this Sequence we're not adding anything together. We're just looking at partial sums. It appears that the sequence is converging To one now we can kind of validate Generically that these kinds of series when the ratio is in a certain category Do in fact converge and we can also categorize the value to which they converge But at least it seems somewhat believable I think with our first example that the sequence of partial sums converges and the sequence converges to one Let me put it in the words of the authors here if the sequence S sub n Remember yesterday. We were talking about things in the curly braces So is it it is a set if the sequence s sub n the sequence of partial sums is convergent then the series itself the one that generated the sequence of partial sums is Convergent and we have a pretty good idea that this particular Series converges to one because the sequence of partial sums gets closer and closer to one now. How can we kind of make a generic description of Geometric series Get a shortcut for the sum of n terms of a geometric series and then a further Shortcut of when they converge when the ratio is in a certain category that is even quicker than If we stop at some specific value, so let's take an infinite actually Let's stop this thing at at a certain number, but let's take an in a geometric series Now I'm going to make the first term a so we can talk about the first term in Doesn't have to say anything about the ratio the ratio is not involved yet So the first term can be anything you want it to be in getting from the first term to the second term We're going to multiply by r and then we're going to multiply by r each term as we go So the second term times r is what? A r times r which would be a r squared the third term notice This is the third term. How many times have we multiplied by the ratio? We've only used the ratio twice even though this is the third term, right? So what's the fourth term of this series? They are cubed is the fourth term So let's go out here to the nth term That's exactly right. So by the time we get to the nth term We will have used the ratio in minus one times and if that doesn't seem quite right Look at the third term the ratios use twice not twice. It's in there squared Fourth term has the ratio cubed. So the nth term We will have multiplied by the ratio in minus one times Used as a used it as a factor in minus one times to get to the nth term So what would be a shortcut of all these terms added together? Just this particular line is not going to help us so what we need to do is Somehow come up with a shortcut that we can see what the sum of n terms in a geometric series would be So let's take this And let's multiply Let me back step a half step. We're going to form a new Equation if this is a legitimate equation We could multiply every term by r and it should also be a legitimate equation So I'm going to take this value on the left side multiply it by r And I'm going to take every value on the right side and multiply it by r. So a times r a r times r a r squared times r a r cubed times r I Left space for this next term If this has the ratio to the n minus first power, how about its predecessor? In minus two right it has used the ratio as a power as a factor one less time As you work your way left. It's always one less Now the reason I did that what is this term times r in minus one? because it'd be a R to the n minus two times r right r is r to the one and When you have product of things with like bases you add the exponents. What's in minus two plus one? Would be in minus one so that's that term times r and Now what's our last term of the original equation? What's a r to the n minus one times r? a r to the n if we added things Everything's going to hang around don't you see a bunch of like terms in the top line in the second line? So if we add them we're not going we're going to keep them around If we see like terms we might ought to want to subtract them if we want them to disappear So let's take this first equation and Let's subtract the second equation from it That looks like a mass The left side is s sub n minus r times s sub n Tell me what is the end result on the right side? What is the top line? Right side of the equal sign Subtract out this set of values down here Close Okay, we've got an a are we going to lose the a with any of these terms so a minus Zero a stays About a r minus a r drops out a r squared minus a r squared a r cubed This guy all these guys knock out with all these guys right and then we're left with what minus a R to the n that helped What we really want is what is s sub n? Do you think we could massage the left side a little bit so we could solve for s sub n? Here's the left side. So here's our equation Good so there's really you're kind of a step ahead there So but I know to get there you had to do this we can on the left side factor out s sub n right? It's in here, and it's also in here How many times is it in the first term one how many times is it in the second term negative r? So let's divide both sides by which was suggested that Jacob. Did you say that okay divide both sides by one minus r? That's a step closer to what is the sum of a specific number of terms in terms in a geometric series You see anything that could be done not that it has to be done with the numerator They both have an a so we could factor an a out of This numerator so that leaves one minus r to the end That is awfully tempting to reduce that numerator with the denominator, but they do not reduce Because this is one minus r and this was one minus r to the end So they're different species not only different animals. They're different species So we're stuck with that, but that's still a pretty good shortcut So let's take the series That served as our first example one over two to the end and started at one and Let's let's go ahead and stop this one Okay, I think we what did I did I use an example where I stopped it I? Didn't so let's just make one up. Let's stop it at five We decided I think that the sum of the first five terms was 31 over 32 And we just added them together and that's not such a bad idea on a problem this simple But they're not all going to be this way and some of them are going to be alternating How could they be alternating by the way? What's that say about the ratio if they're alternating? Positive negative positive negative the ratio must be Negative right so if the ratio is negative Every time you multiply the preceding term by a negative doesn't it change the sign as you progress to the right? So if you've got an alternating series and you think it's geometric you better be looking for a negative ratio This is something you don't have to do. We're just kind of validating it because it's our first Example we're trying to see it through a little further So we know at the last term because we raised two to the fifth, so we use that last value We already know by adding these up that the sum of the first five terms is 31 over 32 now can we get the same thing? Actually, that's s sub five right some of the first five So it should be the first term One minus the ratio to the end. What's the ratio in this series? one half in is five All over one minus the ratio, so we've got a half up here. I'm going to leave that alone here. We've got one over one 32nd one minus a half in the denominator is also half a half over a half knockout one minus one over 32 31 over 32 So not a surprise. We kind of did the thing generically so knew it was going to work and just kind of validated what we had previously Questions on that before we try to take this particular situation just a little bit further Wouldn't that be nice if everything in this chapter were that easy? Oh infinite geometric series They just progress along so nicely and oh it gets a little rough when they're negative because it causes the signs to alternate but No It's not that easy all the way through it Let's take and this is a an example of this which is why I chose this as our first example not just because it's pretty easy Let's take a situation where the absolute value of the ratio is Less than one we know what happens if the ratios One or larger in magnitude don't the terms in magnitude get larger If you multiply by 1.2, don't they get larger as you go If you multiply by negative two even though they alternate signs, don't they get larger in magnitude as you go So those don't have a chance of converging. So if we want this thing to have a chance of converging even though it's infinite Now we're moving on to this Ar to the n minus one That was our nth term Description n starts at one Let's go ahead since we haven't really done it in this closed form before when n is one What do you get you get a because it'd be r to the zero which is one So the first term is a which is what we want it to be if you plug in n equals two into that Description of the nth term. What do you get? they are So that's what we want right first term is a the ratio is r. So we're gaining a factor of r as we go This thing does not stop. However, this one goes on indefinitely, so it is an infinite geometric series and If we are going to use the ratio Smaller than one in magnitude So it's absolute value is less than one. Let's take a look at the sum Well in is headed towards infinity So we're going to have a hard time plugging in r to the end if this thing goes on indefinitely So let's analyze this particular piece if r is Somewhere between one and negative one Let's say I don't know just for first example purposes. Let's say it's a half. We've already used that one So let's continue to use that one What happens to one-half? To the end as n gets infinitely large Doesn't that go to zero Is that correct? One over two one to the end stays one two to the end as n goes to infinity keeps getting larger and larger and larger What happens to a fraction whose numerator is fixed no matter how large and the denominator increases without bound That is zero now would that be the case with any value in this Category I use two-thirds would that also be true? If I use negative three-fourths would that also be true? As long as the numerator is smaller than the smaller than the denominator whether it's positive or negative Eventually that fraction raised to the nth power and n approaching infinity those Always go to zero when r is in this area In that convenient because r to the end if r is somewhere between negative one and one this piece right here disappears Well, we've got one minus zero You know how limited I am, but I can handle that one minus zero is One and the numerator is just A times one which is a that's even easier than this one You would think that if we let it go indefinitely that it would be more complicated. It's actually easier so if it's an infinite geometric series and The absolute value of the ratio is less than one That's a critical piece of this if that's not true then what I'm about to write is not true The sum I'll just use s for some we could subscript it with a little infinity sign, but the sum is A times one minus zero, which is just a over one minus r That's easy Infinite geometric series absolute value the ratio less than one it converges to first term over one minus the ratio So our first example was this One over two to the end that nest doesn't necessarily look like most Geometric series look, but we know it's geometric and we're going to let this thing go on forever. What did we guess? Based on the sequence of partial sums What did we guess that the sum of this infinite geometric series was going to be? Well, that's if we stopped it at five. It was 31 over 32 if we added in the next one it was 63 over 64 if we added in the next term It would be what 127 over 128. What did we think it converged to one? We thought it was going to head to one. Let's see if we can prove that This is really one half plus one fourth plus one eighth Dot dot dot this thing goes on forever If it's an infinite geometric series. Are you convinced of that? It's geometric because the ratio is One half by the way Ratio you normally think of ratio as division, right? How do you come up with the ratio if it's not real clear you take the second term and You divide it by its predecessor Or you take the third term and you divide it by its predecessor That gives you what are it so it is a ratio. It's third divided by second 11th divided by 10th So in all these cases this common ratio is a half and it is infinite So it's an infinite geometric series. Are we in this ballpark right here? Absolute value of the ratio less than one Yes, so this thing should converge to a over one minus r first term is a half One minus the ratio the ratio is also a half a half over a half is One so if we could add these up all the way to infinity we get exactly one And we kind of made an intelligent guess at that based on the sequence of partial sums So it is convergent as soon as we know that it's Infinite geometric and that the absolute value of the ratio is less than one those facts together Tell us that it converges and in this particular case We know the value to which it converges first term over one minus ratio questions on that before we Go just another half step ahead that doesn't seem like it fits repeating decimal But it does fit because a repeating decimal is an infinite geometric series Any decimal let's dig back into our mathematical background. That's always fun to do If a decimal repeats or terminates what kind of number is that I told you it was fine necessarily fun for you The decimal repeats that this is a repeating decimal or it terminates point two five That's a terminating decimal That is a what kind of number? Rational number very good So let's continue back in the memory banks a Rational number can always be written. I mean normally when you think of a rational number you think of a fraction Right a fraction is a rational number So any rational number first of all it's any decimal that repeats or terminates If it doesn't repeat or doesn't terminate what kind of number do we call that? Ear rational not rational. Okay doesn't repeat doesn't terminate But these this is repeating Even if it terminates doesn't it repeat in zeros? So in essence they're all repeating in this category. It's any number that can be written in the form of a over b See if this sounds familiar where a and b are integers and b is Not equal to zero That's probably the first definition you had Whenever you had that seventh grade eighth grade the first definition of a rational number any number that can be written in the form of a over b Where a and b are integers and b is not equal to zero. It's a fraction rational number is a fraction So this is a rational number So we should be able to write this as a fraction Not just a fraction a fraction of two integers And the denominator is not going to be zero obviously The first point three seven Isn't that 37 over a hundred correct the next 37 we encounter is really point zero zero three seven because we've already taken care of the first two decimal places What is point zero zero three seven? 37 over Two more zeros right then that's four zeros 10,000 the next 37 we encounter would be 37 over Whatever that is two more zeros and so on right so aren't these terms added together the same as 37 over a hundred plus 37 over 10,000 plus 37 over one with six zeros Those are the same thing and in fact aren't they this guy right here? They are point three seven repeating decimal So is it a series? Yes, it's a series. It's infinite because this thing repeats through infinity and What is the ratio? Well, let's take a half step back. Isn't that the first term? 37 over a hundred is the first term of the series What do you multiply by to get from the first term to the second term? one over 1,000 or 100 we got two more decimal places to accommodate so 100 so the ratio is One over 100 if it repeats in blocks of two it has a ratio of one one hundred Soon we be able to find the sum it's an infinite geometric series So the sum should be first term Which is thirty seven one hundredths over one minus the ratio So that's thirty seven Over a hundred one minus one one hundred is ninety nine over a hundred So can we write this rational number? We better be able to if it's a rational number as the quotient of two integers We certainly can and it would be what? Thirty seven over ninety nine if you don't believe that's thirty seven thirty seven thirty seven repeating decimals Take your calculator right now and take thirty seven and divide it by ninety nine Work right point three seven repeating So what's that written as a fraction? Fourteen over ninety nine right for repeats in blocks of two like this just like the previous one Check it out fourteen over ninety nine. Let's just do a little bit of off-the-side pattern analysis here This is easy Point three it repeats in blocks of one so the ratio would be instead of one one hundredth. It would be one tenth So it'd be first term over one minus the ratio Three tenths over nine tenths not a mystery that this is three ninths right So if it repeats in blocks of one it's ninths, you know where I'm headed with that one our old friend point nine repeating decimal Shouldn't it be the first term Over one minus the ratio. Oh my goodness It's one it's nine ninths there it is again. It just kind of keeps rearing its Ugly head. I don't know. I think it's kind of pretty myself Point nine repeating decimal is actually one All right, let's go the other way. Let's say it's 281 and it repeats in blocks of three 281 Over 999 Check it out take 281 divide it by 999 All of this stuff just comes from the fact that if it is an infinite geometric series The son and the ratio by the way is less than one in absolute value, which it is It's a over one minus R now the only other kind of problem that you might encounter other than Blocks of one blocks of two blocks of three would be where part of it is stable and part of its repeating So let's say we have a four tenths kind of out there by itself and the repeating stuff Doesn't really start till we get to the one in the eight So you do want to cast that non repeating part aside And we'll add that in at the end of the problem and then deal with the part that repeats Well the part that repeats is slightly different from the way we looked at it earlier. The first term is 18 What? 1,000th, right? It's not in the first two decimal places the first repeating block of two Is actually covering the first three decimal places Then once we take those and get rid of them Then this three zeroes now in the bottom gets increased by two, right? Because the ratio is now in blocks of two. It's one one hundredth as we progress so eighteen over a Thousand is our first term Over one minus the ratio the ratio is one one hundred One minus one one hundredth is ninety nine over a hundred Not quite as easy of a fraction where we could just you know say that this one's over a hundred This one's over a hundred multiply numerator and denominator by a hundred But we could multiply numerator and denominator by a thousand so we should get 18 over 990 Somebody with your calculator out check that out. What is 18 over 990 set this Point zero one eight one eight Got that We're not completely done. We've got to add four tenths to that So we've got to combine those two fractions. What can we do to the first fraction four over ten? So that we can combine it with 18 over 990 That's about 99 multiply by 99 99 numerator and denominator So now we've got nine ninetieths in each fraction. What's 99 times four plus 18? four 14 is that right So our final check would be to take a calculator and take four fourteen and divide it by 990 And if we've done everything correctly, we should be back to point four one eight one eight That work so repeating decimals are infinite geometric series Okay, we're not always given the series in the kind of a closed Sigma format. We might just have it in expanded form Did any of you go to see John Pizzarelli last night? Were any of you there at the? Stuart theater jazz guitarist and singer Nobody was there man. You missed it. It was so good John Pizzarelli It was it was excellent Yeah, but if you get a chance to ever listen to that person if you like jazz kind of an old-timer music like I would listen to from time to time He was unbelievable Sorry if that's an aside What kind of series is this? If you think it's geometric, I guess that's one of the reasons why I wanted this to be an example What should you do if you think it's geometric? Second term divided by first term and is that equal to let's say fourth term divided by third term is it what is that ratio? What are we multiplying by? Negative one third is that correct and you can check it out two times negative a third is this This term times negative a third is this this term times negative a third is this so Doesn't take long a few seconds to kind of validate that visually So it is geometric The dots that I see tell me it's infinite Geometric so if you are asked on a test is this a convergent series and if it's convergent What value does it converge to? This is your justification. It's an infinite geometric series. What else do we need to say the ratio is negative one third and Because we know the ratio is negative a third The absolute value of the ratio is less than one if that's not true. It doesn't converge Absolute value of negative a third is a third which is clearly less than one Therefore it's convergent and can't we go a step further and find the value to which it converges Yes, we can and it is first term over one minus the ratio first term is two ratio is negative one third Remember that we're subtracting the ratio in this case. We're subtracting a negative That's kind of hard to figure out by the way by looking up here because they alternate You've got to and then you subtract two-thirds and then you add two-ninths and you subtract two-twenty-sevenths. That's not necessarily visually easy to figure out so it's better just to go right to this That's one plus one-third, which is four-thirds So if we could add all of these terms together all the way out to infinity what would be the sum? Three-halves Which is two times? Three-fourths, which is three-halves. I would have never guessed three-halves by Looking at these four five or eleven terms that we happen to write out initially so this Can be trusted as the sum of an infinite number of terms in a geometric series as long as the ratio is in this Area right here. All right good. So this is going to be perfect We have one more thing with infinite geometric series to look at and then we'll be ready to look at what are called Telescoping series and we'll start that tomorrow. Okay. This is slightly different because Well a couple of reasons it's got an x in the argument that talks about what we're going to substitute into n starts at zero There's our other difference Most of the time it's common to start them at one because it's very handy to let n equal one Generate the first term Kind of seems logical n equals two generates the second term n equals three generates the third term and so on It's kind of a little bit behind if you think of n equals zero generating the first term But whatever we're handed here, that's where we're supposed to start So what would this thing look like if we wrote it out? first term is one Second term is x to the n and n is one so x to the one is x When n is to x to the two is that the way that series goes Is that a geometric series? We're multiplying by R and R is x so we multiply by x Multiply by x Multiply by x and I know we don't know what x is but it is x the ratio is x it is an infinite geometric series If it converges and we've got a little big if there If it converges what would have to be true in order for this to converge? Absolute value x has to be less than one we don't know that because we don't know what x is but if x were in this area Then it converges and if x in absolute value is less than one This thing converges to this Which what would that be in this particular problem? So it may not look right to you, but one over one minus x is Equal to one plus x plus x squared plus x cubed So there's an infinite geometric series that represents this pretty simple fraction And I'll put a little disclaimer here if The absolute value the ratio which is x in this case is less than one Perfect timing. We are a 10-05 minus about three seconds 11-05 so I will see you tomorrow