WEBVTT mathematics/calculus-ii/murray
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Hi, this is educator.com and this is the chapter on sequences.
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We are going to be exploring some different ways to find limits of sequences.
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There are several definitions that lead us up to a big theorem that sometimes be a very powerful way to show that a sequence converges.
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Also, to show its limit.
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The definitions we have are monotonically increasing means that the terms of the sequence are getting steadily bigger.
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The a < font size="-6" > n < /font > term is getting steadily bigger in other words the a < font size="-6" > n+1 < /font > term is bigger than or equal to the a < font size="-6" > n < /font > term for all n.
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Monotonically decreasing, same idea except the terms are getting steadily smaller.
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The word monotonic just means the sequence is either monotonically increasing, or monotonically decreasing.
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If either one of those is true then you say the sequence is monotonic.
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Bounded means that all of the terms of the sequence means that all of the terms in absolute value are less than some constant number m.
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The big theorem that we are going to use here is that if you can show that a sequence is both bounded and monotonic,
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Meaning that if you can show it is monotonically increasing or decreasing, and that is bounded,
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Then the sequence converges.
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Let us see how that works with an example.
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The example here is a sequence that starts at the sqrt(2).
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Then the way you get successive terms is you take a previous term is by adding 2 and taking its square root.
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For example, a₁ would be the sqrt(2+a₀), so 2 + sqrt(2)
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a₁ would be the sqrt(2+a₁), and so on.
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We have to show that that sequence converges and then we are going to find its limit.
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We will use our definitions and theorem.
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Our first claim here is that the sequence is bounded above by 2.
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In other words, I claim that every term in the sequence is < or = 2.
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To prove this, note that the first term of the sequence, the a a₀ term,
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Is certainly less than 2 because the sqrt(2) is about 1.4.
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That is certainly less than 2.
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If a < font size="-6" > n < /font > is < 2, if one term < 2,
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Then I am going to a do a little arithmetic here.
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2 + a < font size="-6" > n < /font > would then be less than 4.
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So the sqrt(2+a < font size="-6" > n < /font > ) would be less than sqrt(4) which is 2.
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That is saying that a < font size="-6" > n < /font > + 1 < 2.
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If one term is less than term, then the next term is less than 2.
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We already showed that the first term was less than 2.
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So, every term is less than 2.
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Each term forces the next term to be less than 2.
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That proves the claim that the sequence is bounded above by 2.
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That really is an argument by mathematical induction.
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The second claim is that the sequence is monotonic.
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I claim that this sequence is monotonically increasing.
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So I claim that each term is bigger or equal to the previous term.
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The check whether this claim is true, this is true if and only if
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Well to be monotonically increasing, is to say that a < font size="-6" > n < /font > + 1 is bigger than or equal to a < font size="-6" > n < /font > .
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Let us plug in what a < font size="-6" > n < /font > + 1 is.
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By definition that is 2 + sqrt(2 + a < font size="-6" > n < /font > ), bigger than or equal to a < font size="-6" > n < /font > .
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I am going to work with this a little bit.
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I am going to square both sides, this is saying 2 + a < font size="-6" > n < /font > is bigger than or equal to a < font size="-6" > n < /font > ².
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If I move all of the terms over to the right, that is saying 0 is bigger than or equal to a < font size="-6" > n < /font > ² - a < font size="-6" > n < /font > - 2.
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I can factor that into (a < font size="-6" > n < /font > - 2) × (a < font size="-6" > n < /font > + 1).
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Let us remember that since a < font size="-6" > n < /font > < 2.
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Over on the left we showed that every term is < 2.
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Since a < font size="-6" > n < /font > < 2, a < font size="-6" > n < /font > - 2 < 0.
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Since all of these terms are positive, a < font size="-6" > n < /font > + 1 would be bigger than 0.
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a < font size="-6" > n < /font > - 2 × a < font size="-6" > n < /font > + 1 is less than or equal to 0.
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Which means that this equation is true and this equation is true.
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a < font size="-6" > n < /font > + 1 is > or = a < font size="-6" > n < /font > .
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What we showed here is that the sequence is bounded and monotonically increasing.
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Our theorem kicks in here and says a bounded monotonic series converges.
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We can invoke the theorem.
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By the theorem the sequence converges.
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That shows the sequence converges, so we have done half of the problem there.
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The sequence converges.
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The second half is to find its limit.
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We will do that on the next page.
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To find its limit, we know now that it does have a limit because we showed that it converges on the previous page.
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Let us call that limit L.
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That is saying that a < font size="-6" > n < /font > converges to L.
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There is a little trick here which is that if a < font size="-6" > n < /font > converges to L, then the sequence of successive terms must also converge to L.
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But a < font size="-6" > n < /font > + 1 is the sqrt(2 + a < font size="-6" > n < /font > )
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So, a < font size="-6" > n < /font > + 1 = sqrt(2 + a < font size="-6" > n < /font > ).
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But, a < font size="-6" > n < /font > converges to L, so sqrt(2 + a < font size="-6" > n < /font > ) converges to 2 + L.
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Now we have a < font size="-6" > n < /font > converging to L, but it also converges to 2 + L.
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So sqrt(2 + L), L must be equal to sqrt(2 + L).
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Now we want to solve that equation and figure out what L is.
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L² = 2 + L, L² - L - 2 = 0.
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(L-2)(L+1) = 0.
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L is either 2, or -1.
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The limit is 1 of those two possible numbers.
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Remember that all of these terms are positive
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So, you cannot have a bunch of positive numbers converging to -1.
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We cannot have a < font size="-6" > n < /font > converging to -1, so a < font size="-6" > n < /font > must converge to the other possible limit, 2.
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Let us recap that problem there.
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We were given this fairly tricky sequence and we had to show that it converges first
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The way we showed it converges was we tried to show that it was bounded and we showed that it was monotonic.
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Monotonically increasing.
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Then those two together allowed us to invoke the theorem which says a bounded monotonic sequence converges.
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Those two conditions allowed us to say that the sequence converges.
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Then once we know it converges we can assume that its limit is L and go through this little algebraic trick looking at what a < font size="-6" > n < /font > + 1 converges to.
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Set those two limits equal to each other because the sequence only converges to one limit.
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Do a little algebra to find the possible limits.
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Then see which limit makes sense.
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The limit that makes sense was 2 and so the limit must be 2.
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Let us try something a little more computational.
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Here we are given the sequence 3n + 5n²/2n² + 6.
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The trick here is to find the largest term in the numerator and the denominator and divide.
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Here we look at the numerator 3n + 5n², definitely the largest term there is 5n².
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In the denominator, 2n² + 6, even though the 6 is bigger than the 2, the n² is going to be much bigger in the long run.
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So the 2n is going to be much bigger there.
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In fact, what really matters here is not the coefficients, it is the powers on the n's.
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The fact that we have an n² in the numerator and the denominator is the important thing.
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What we are going to do is divide top and bottom by n².
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When you divide top and bottom by the same number, that is really multiplying all of it by 1.
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That is legitimate and it is not going to change the limit.
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We get (3n/n² + 5n²/n²)/(2n²/n² + 6/n²).
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(3n/n² is 3/n + 5)/(2 + 6/n²).
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In the limit, as n goes to infinity, the 3/n is going to go to 0.
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The 6/n² is also going to go to 0 so we end up with 5/2 as our limit.
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To recap when you have one of these fractional situations
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Where you have an expression in the numerator and an expression in the denominator.
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You want to look at both numerator and denominator and find the largest term in sight,
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And divide both top and bottom by that largest term.
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Here the largest term was n², so we divide top and bottom by n².
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That really makes all the other terms go to 0, and it leaves you with the terms that you need to determine the limit.
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Let us try another limit example.
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We want to find the limit of the sequence of sqrt(n)/ln(n).
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Here, both of these sqrt(n) and ln(n), when n goes to infinity, both of these go to infinity.
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What we are really looking at is a situation of something going to infinity/something going to infinity.
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The classic rule to use here that you might remember from Calculus 1 lectures is l'Hopital's Rule.
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Remember, you can use l'Hopital's Rule in two situations.
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You can use it when you have a limit going to infinity/infinity or 0/0.
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If you have 0 × infinity, you cannot use l'Hopital's Rule directly,
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But you try to rewrite it as an infinity over infinity or 0 over 0 situation.
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Then you can use l'Hopital's Rule.
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Let us remember what l'Hopital's Rule does.
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It says you can take the derivative of the top and bottom.
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The strange thing here is you do not take the derivative using the quotient rule you learned in Calculus 1.
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Instead you take the derivative of the top and bottom separately.
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The derivative of the sqrt(n), we think of that as being n^1/2.
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Its derivative is 1/2n^-1/2, that is square root of n in the denominator.
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The derivative of ln(n) is 1/n.
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If we take the denominator and flip it up to the numerator, we get n/2sqrt(n).
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n/sqrt(n) is the sqrt(n).
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Now we want to take the limit as n goes to infinity of sqrt(n/2).
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That clearly goes to infinity.
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We say that this sequence diverges to positive infinity.
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To recap what we did with that example, we were given sqrt(n)/ln(n).
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We noticed both of those go to infinity, so we have a limit that goes to infinity over a limit that goes to infinity.
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That is l'Hopital's Rule.
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l'Hopital's Rule says you take the derivative of the top and bottom and you take those two derivatives separately.
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You do not use the regular quotient rule that you use in Calculus 1.
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We simplify it using a little bit of algebra, and then we talk the limit of the new expression that we get.
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Then that tells us what happens to the original sequence, it diverges to infinity.
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Let us try some more examples later.
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