WEBVTT mathematics/calculus-ii/murray
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Hi this is educator.com
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We are going to talk today about series
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There are several bits of notations and definitions before we look at some examples
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The first is this big sigma notation.
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The notation this symbol sigma stands for the series where you plug in different values of n.
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This is just short hand a₀, a₀ + 1, a₀ + 2.
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The way you want to think about these series is by thinking about the sequence of partial sums.
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What you do is you add up these series 1 term at a time.
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First you start with the 0 term if there is one.
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Then the second partial sum is a₀ + a₁,the third is a₀ + a₁ + a₂ and so on.
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The s < font size="-6" > n < /font > is a₀ + a₁ up to a < font size="-6" > n < /font >
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You call this the sequence of partial sums.
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s₀, s₁, s₂ up to s < font size="-6" > n < /font > .
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You think of that as a sequence, not a series.
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This is a sequence s < font size="-6" > n < /font > .
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We want to say what it means for a series to converge.
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You have to be quite careful when you make this definition.
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It is not quite as obvious as you think.
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What you say is you look at the sequence of partial sums.
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We view that as a sequence
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Then we go back to our definition for a sequence converging.
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if the sequence of partial sums converges, to a particular limit.
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Then we say the series converges to that limit.
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By definition, a series converging means the sequence of partial sums converges.
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In several of the examples that we will see later on,
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We will be given a series and the way we will handle it is we will write out the partial sums and then we will think of them as the sequence.
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We will see what happens to that sequence.
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The same thing holds for all the other possibilities.
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Diverging, diverging to infinity, or diverging to negative infinity.
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You look at what the sequence of partial sums does.
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Whatever behavior that does, you say the series does the same thing.
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There is a very common type of thing that you will see.
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We call it a geometric series.
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It is one where each term is equal to the previous term multiplied by the same common ratio.
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In practice that looks like a + some a × r + a × r² and so on.
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The important thing is that each term is getting multiplied by the same number every time.
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We have a formula for the sum of a geometric series.
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What you have to do is determine first of all if the ratio and absolute value < 1 or > 1 or = 1.
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If the ratio and absolute value < 1, the series adds up to a/1-r.
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This is the formula you will see in a lot of books.
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I think this formula can be a bit misleading.
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Because it depends on whether you start the series at n = 0, or at n = 1.
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Some books have the a/1-r formula, some books have a slightly different formula.
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It depends on whether they are using n = 1 or n = 0 as the first term in the series.
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I do not like that formula so, I will give you a fail safe formula that will work in all situations right here.
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First term, divided by 1 - the common ratio of the series.
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That formula always works in all geometric series.
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When your ratio is < 1.
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I think that is the one to remember, even though it is words instead of an equation.
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That is the one that will get you through any geometric series.
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If the common ratio is bigger than 1 in absolute value
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Or if it is equal to -1, than the series just diverges.
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If the common ratio, if r = 1, then that means that what you are doing is you are adding a + a + a.
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That clearly diverges either to infinity, if a > 0 , or -infinity, if a is negative.
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You will probably not get geometric series with r - 0 because they are too simple to be given in a calculus exercise.
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One more theorem that we are going to be using is called the test for divergence.
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It is usually the first thing that you want to test with every series.
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If you are given a series, what you do is you look at the individual terms.
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You see whether the individual terms converge.
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If they converge you ask what do they converge to.
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If they converge to anything other than 0, then you can immediately say the series diverges.
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Also, if the sequence of terms diverges, you can say the series diverges.
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To recap, you look at the individual terms, if they converge to something other than 0, or if they diverge.
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Immediately you can say the series diverges.
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The flip side that often mixes up students.
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If the sequence of terms does converge to 0, people think that you can use that to conclude that the series converges.
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That is not true and we will see examples of that later on.
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Then just from that information you cannot conclude anything about the series.
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You have to go and find one of the other tests that we will discuss later on.
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If the sequence converges to 0, you are really stuck.
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You cannot use the test for divergence, however, if the sequence converges to something other than 0,
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You can use the test for divergence immediately to say that the series diverges.
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To emphasize here, the test for divergence can tell you that a series diverges, but it can never tell you that a series converges.
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This is a common mistake made by students.
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People will say oh a series converges by the test for divergence.
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That is a very bad mis-use and your teachers will have no patience with that.
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You can use the test for divergence to say that a series diverges,
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But if you get that the series converges, if the sequence converges to 0, the test for divergence tells you nothing and you have to find something else.
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Let us try some examples.
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First example here is the series n-1/n
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Right away we will look at the test for divergence and see if it works.
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a < font size="-6" > n < /font > = n-1/n.
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We rewrite that as 1-1/n.
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The limit of the sequence a < font size="-6" > n < /font > is well the 1/n will go to 0 is 1.
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This is not 0.
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So the sequence of terms converges to something other than 0.
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The sequence a < font size="-6" > n < /font > converges to something other than 0.
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The series diverges by the test for divergence.
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The test for divergence says you look at the sequence of terms, see if they converge to something other than 0.
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If so, then the whole series diverges.
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If this had come out to be 0, if the limit had been 0, then we would not know and we would have to go on and find some other test there.
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Let us try another example here.
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The series of 1/n so again you will look at the test for divergence.
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a < font size="-6" > n < /font > = 1/n and the limit of that as n goes to infinity is 0.
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The sequence converges to 0.
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The test for divergence, if the sequence converges to 0, tells us nothing.
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So, t(d) fails to give us an answer here.
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We cannot say anything yet.
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Instead we will look at the partial sums.
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s₁ is just the first partial sum, that is just 1.
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s₂ is 1 + 1/2.
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s₃ is 1 + 1/2 + 1/3.
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s₄ is 1 + 1/2 + 1/3 + 1/4.
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I am going to start adding these numbers up and see what kind of sums we get.
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Just look at 1, it sounds kind of silly to say it, but 1 > or = 1.
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We will see why I am making a big deal out of that later on.
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1 + 1/2 > or = 3/2, in fact it is equal to 3/2.
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I will skip s₃, I will not look at that.
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s₄, if you look at 1/3 + 1/4 now 1/3 is bigger than 1/4.
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So this is > or = to 1/4 + 1/4.
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What we have here is 1 + 1/2.
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Plus something bigger than 1/2.
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This is bigger than 2.
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Now I am going to start skipping, I am going to look at s < font size="-6" > n < /font > .
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s < font size="-6" > n < /font > is 1 + 1/2 + 1/3 + 1/4 + 1/5 all the way up to 1/8.
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Again, we have 1, that is bigger than 1.
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1/2 is at least 1/2.
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1/3 + 1/4 is bigger than or equal to 1/2.
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Here we have 4 terms, each one of those is bigger than or equal to 1/8.
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Collectively they are bigger than or equal to 1/2.
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What we have here is 1 + 1/2 + 1/2 + 1/2.
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This whole thing is < or = to 5/2.
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Now I will skip to s < font size="-6" > 16 < /font > .
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Without showing all the individual terms, it is bigger than 1/2 + 1/2 + 1/2 + 1/2 and then we will have 8 more terms, all bigger than 1/16.
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So 1/2, this is bigger than or equal to 3.
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If you look at the sequence of partial sums as a sequence, we have got 1, 3/2, 2, 5/2, 3.
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Clearly, the sequence of partial sums s < font size="-6" > n < /font > diverges to infinity.
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By definition, the series 1/n diverges to positive infinity also.
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There are a couple of points we want to make to recap that.
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We tried to use the test for divergence.
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That says you use the sequence for individual terms, but those went to 0, so the test for divergence tells us nothing.
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Instead, we look at the sequence of partial sums.
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That means we start adding up these terms and we kind of group them together in clever ways.
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When we group them together in clever ways, we notice that the sequence of partial sums is going to infinity.
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So, we say the entire series diverges to infinity.
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A couple more things that I want to say about this series.
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One is that it is very well known, it is called the harmonic series.
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It is called the harmonic series because it arises in looking at musical notes.
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This is well known and is called the harmonic series.
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People will tell you the harmonic series diverges and the reason is by this proof here.
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Another important thing to note about this series is we could write this as the sum of 1-n^p where p=1.
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It is really 1/n¹.
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This is a special case of something we are going to see later called the p series.
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That is kind of a preview of something we will see later when we talk about the integral test.
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We will be talking about p series.
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The harmonic series is a special case of p series, with p=1.
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Let us try another example here.
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We are trying to determine if the series 1/n+1 converges or diverges.
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Again, the trick here is to look at the partial sums and before we right out the sequence of partial sums.
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We are going to do a little algebra here.
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We are going to try to use partial fractions on 1/n × n+2.
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Partial sums is an algebraic trick that we learned back in the partial sums section.
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It says we can separate 1/n × n+2.
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It says we can separate a/n + b/(n+2).
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Then we can try and solve for constants a and b.
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We learned this in detail in the lecture on partial fractions.
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If you are a little fuzzy on partial fractions, we might want to go back and review that lecture,
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In the meantime, I am not going to work it out but I am going to tell you the answer is a=1/2 and b=-1/2.
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That comes from applying partial fractions here.
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There is a little bit of algebra and solving 2 equations and 2 unknowns that I am suppressing here.
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But, you can work it out and get these values for a and b.
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We can write this as 1/2/n - 1/2/n+2.
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If I factor out the 1/2 there, I get 1/2 × 1/n - 1/n+2.
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That is going to be very useful in trying to figure out what the partial sums are.
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Let me write down what a few of those partial sums are now.
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s₁, plug n=1 in here and we get 1/2 × 1 - n=1, gives us 1/3.
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I could simplify that but I am not going to because later on it will be easier to spot a pattern if we do not simplify that.
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s₂ is 1/2 × 1 - 1/3, + the n = 2 term, is 1/2 - 1/4.
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s₃ is 1/2 × 1 - 1/2 + 1/2 × 1/2 - 1/4 + 1/2 × the third term is 1/3 - 1/5.
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Now you start to notice some cancellation here.
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This 1/3 will cancel with that 1/3, which is why I did not want to simplify earlier.
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Let me write out 1 more term, s₄.
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1/2 × 1/3 + 1/2 × 1/2 - 1/4 + 1/2 × 1/3 - 1/5 + 1/2 × 1/4 - 1/6.
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Now you start to see a lot of cancellation.
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This 1/3 cancels with this 1/3, this 1/4 cancels with this 1/4.
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If there were another term, the next term would have a 1/5, and that would cancel with that 1/5.
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So, it looks like there is going to be a lot of cancellation as we write out more partial sums.
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Let me try to write a general term for s < font size="-6" > n < /font > .
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It is going to be 1/2, well after we cancel everything, the only terms left are this 1
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That never cancels, the 1/2 never cancels.
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Then the very last terms do not cancel, but everything in the middle is all cancelled.
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We get 1/2 × 1 + 1/2 -, well the last two terms.
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When n=4, the last 2 terms were 1/5 and 1/6.
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Those last two terms are n+1 and 1/n+2.
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So, this simplifies down to 1/2 × 3/2 - 1/n+1 -1/n+2.
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The limit as n goes to infinity of the partial sums.
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Remember we think of the partial sums as a sequence now.
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Those terms just go to 0.
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We are left with 1/2 × 3/2 = 3/4.
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We have the sequence of partial sums going to 3/4.
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We say the series converges to 3/4.
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To recap there, when you are given a series,
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You want to determine whether it converges or diverges.
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That depends on what the sequence of partial sums does.
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That is s₁, s₂, s₃, s₄.
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What we did with this particular one is we did kind of some algebraic cleverness.
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In breaking n/n+2 up using partial fractions.
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That gave us an expression that when we wrote out the partial sums,
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They all cancelled in the middle and just left us with these beginning terms, and these ending terms.
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This happens often and it is called a telescoping series.
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It is called a telescoping series when you write out the sequence of partial sums and the middle terms all cancel,
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Leaving you with just the terms at the beginning and the end.
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Once you simplify it down to something in terms of the beginning and the end,
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You can take the limit, whatever limit you get is, the limit of the sequence of partial sums.
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By definition that is also the limit of the series.
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We will try some more examples later.
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