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Today we're going to be talking about how to determine whether or not the sequence converges.



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And if it does converge how to find its limit.



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And in this particular problem what we've been given the sequence A7 is equal to cosign squared of N



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divided by 2 to the end power.



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Now before we start in on this problem I want to talk a little bit about the instructions of the problem.



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These are the instructions that we've been given for this particular problem but I feel like they're


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a little bit misleading and often we're asked to find whether or not a sequence converges and then if



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it does to find its limit but that can be a little misleading because when it comes to sequences and



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keep in mind that this has nothing to do at all with series series are completely separate.



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This is for sequences only for sequences only.



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We can almost reverse these directions these instructions because if we can find the limit of a sequence


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and that limit exists then by definition the sequence converges.



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So these instructions could just as easily say find the limit of the sequence if it exists and then



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use your answer to determine whether or not the sequence converges So that's what we're going to do.



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I'm also going to show you how to use squeeze theorem quickly to determine first whether or not the



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sequence converges.



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And then to draw a conclusion about the limit.



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If you feel more comfortable doing it that way.


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But the first thing we're going to do here is try to find the limit.



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So we're talking about the limit as and approaches infinity of our sequence of the limit as it approaches



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infinity of a sob in that case of course is always just equal to the limit as and goes to infinity of



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our sequence here.



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Co-sign squared of n over 2 to the end power.



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Now the easiest way to try to find the limit when you have a rational function like this is to deal


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with the numerator and denominator separately.



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So if we look at the numerator here first cosigners squared of n and we're trying to figure out what



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the value is of this as and goes to infinity.



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We need to think about the cosigned function.



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Remember that the coastline function oscillates back and forth between the values negative 1 and positive



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one we know that from the graph of cosign.


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We also know it from the unit circle that the smallest value cosign ever attained is negative 1 the



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largest value it attains is positive ones so we can almost say cosign of N is going to be greater than



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or equal to negative 1 and less than or equal to positive one.



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So we already know that.



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Now if we square the cosign function right because we have cosigned squared of and here if we square



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the cosigned function all the values of cosigned between negative 1 and 0.


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All those negative values when we square them they become positive.



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But they're all going to be positive values between zero and positive one.



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Right.



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If we square the left limit here negative 1 we get positive 1 if we square zero we just get zero.



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And everything in between all the way from negative 1 to zero.



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All those you know negative points by values we square those they become positive values between 0 and



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1.


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So what we can say is that cosigners squared of N on the other hand will always be less than or equal



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to 1 but greater than or equal to zero.



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All the values now are going to fall in between 0 and 1.



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So that's our conclusion about cosigned squared of N that this numerator here is just going to oscillate



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back and forth between 0 and 1 and I'll never be less than zero.



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It'll never be greater than 1.


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So that's going to be some constant value maybe we can call it k we'll say the limit as an goes to infinity



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of K because it's always going to be between 0 and 1.



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Now we have to deal with two and here.



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So as we take the limit is and goes to infinity of 2 to the n.



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You can almost think about plugging in infinity here for and we have to do the Infinity well.



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Either way the larger you make this exponent the larger the denominator gets right 2 squared is four.



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But two to five is 32.


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And that are you just going to keep getting larger and larger as we take and larger and larger so the



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limit as and approaches infinity of this to the end here is going to be an infinitely large number so



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we can almost just think about it as k over infinity some huge number.



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Well whenever that's the case this is just going to be some value between 0 and 1.



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Either way the smallest value we would ever have here is zero over infinity.



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Right.


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And that's obviously just going to be 0 or the largest value would have for k would be 1.



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And when we divide that by infinity that's also going to give us zero because when we take any constant



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number and we divide it by an infinitely large number as this becomes larger and larger and larger This



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is always going to go to zero.



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So what this tells us is that the limit as end goes to infinity of our sequence here that this is going


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to be 0 that the limit as and goes to in the are sequence a Sabun is going to be equal to zero.



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This tells us that the limit exists that the limit is zero.



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And because the limit exists and it's zero we automatically know therefore that the sequence converges.



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And so notice that we instead of following the instructions in the order that they gave them to us we



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first found the limit.



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We proved the limit was equal to zero.


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We found that it existed and because it existed we know by the definition of convergence for a sequence



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that the sequence does in fact converge.



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So you can just find it that way if you want to try to prove without the limit necessarily that the



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sequence converges.



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You can always try to use something like squeeze theorem and it would look something like this that



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we already started right where we said that cosign of and was always going to be greater than equal


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to negative 1 and less than or equal to 1 we squared cosign again.



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Now if we multiply this through by one over two to the end because we're trying to get this middle function



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here back to our original sequence.



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So when you multiply through one or two to the end times zero is still zero.



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The middle function here becomes cosign squared of an over 2 to the n r sequence and then less than



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or equal to we just get 1 over 2 to the end.


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Now if we take the limit of this whole inequality as and goes to infinity we take the limit of the whole



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thing.



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What we can do is distribute this limit across at each one so the limit as end goes to infinity of zero



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is still just 0.



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OK so less than or equal to than the limit as an goes to infinity of our function here cosign squared



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of an over to the N or sequence less than or equal to the limit as and NGOs to infinity of 1 over 2



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to the end.


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Well we already know from looking at this equation here that when we have some constant divided by some



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infinitely large number when we plug in infinity here for n this denominator is going to get infinitely



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large.



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We know that this value right here is going to go to zero that's going to become zero.



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So what this tells us is that the limit as and goes to infinity of our sequence because it's always


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going to be greater than or equal to zero and less than or equal to zero by the squeeze theorem.



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We've proven that this converges to zero.



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That the limit is equal to zero.



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So that's another way.



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Without first finding the limit.



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Indirectly that's another way to prove that the sequence converges.



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And then to find that the limit is also equal to zero by the squeeze them.