Original subtitles

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Today we're going to be talking about how to estimate the error or remainder of the series using the

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first 10 terms of the series.

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And in this particular problem we've been given the infinite sum from any equals 1 to infinity of 1

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divided by the square root of the quantity and to the fourth plus one.

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One thing that I quickly want to point out error versus remainder of the series.

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It's really the same thing.

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Some books and professors will call it the air.

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Others will call it the remainder.

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But it means the same thing and you can be asked either way whenever you're asked to find the air or

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remainder of the series you want to do.

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Three things really.

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First of all you want to identify how many terms you're going to use.

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And in this particular problem we've been given that figure exactly we've been told that we're going

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to use the first 10 terms of the series.

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What you want to do with the first 10 terms is find the some of the first 10 terms and then estimate

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the remainder of the first 10 terms and the remainder or the air is going to be different depending

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on how many terms you use so it's really important to know how many terms of the series we're using

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and in this case we're using 10.

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So the first thing we'll do is find the some of the first 10 terms then we'll use that to find error.

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So the first thing we're going to do is find the some of the first 10 terms and then after that we'll

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find air.

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So finding some is really easy the first thing we want to do is just identify that this value here one

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divided by the square root of end of the fourth plus one is equal to a sub and we'll just call a sub

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in our series that'll represent our series.

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So when we want to find the sum to the first 10 terms we'll pretend that the sum of the series is Subhan

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and therefore finding the sum of the first 10 terms is going to give us as sub 10.

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So we'll call it as subtend.

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And really this is just going to be plugging in the numbers 1 2 three four five six seven eight nine

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10 for N and adding all those values together.

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So we'll plug in one for and one to the fourth is just one plus one is two.

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So the first term is 1 over square root 2 then only plug into.

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We get 2 to the fourth which is 16 plus 1 which is 17.

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So that's 1 over square root 17.

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If we plug in 3 3 to the fourth is 81 plus one is 82 so we get 1 over the square root of eighty two.

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And we keep going like that all the way until 10 and when we get to 10 we can just say dot dot dot here

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when we get to 10.

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We'll have the square root 1 over the square root of the fourth is ten thousand plus one is 10000 ones

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will say 10000 and one we take the square root of that.

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So you need to find all the ones in the middle but then you add them together and you can do this on

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your calculator and it's perfectly acceptable to provide a decimal representation of the sum.

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So in our case the decimal value or the value of the sum of the first 10 terms rounded to four decimal

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places is 1 point 2 4 8 6.

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Now when it comes to estimating the error or remainder what we want to do is use the comparison test

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to verify that a 7 converges and then use the remainder of the comparison Series B 7 to more specifically

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define the remainder of our given series a 7.

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So when we're looking for a comparison series for our series we want to do is pick a series that is

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similar to ours for which we can easily determine convergence.

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And that takes a little bit of practice but one thing you can look for are values in your series that

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are kind of trivial to the sum and what I mean by that is for example.

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To the fourth is going to have a much greater effect on this series than this value here is this plus

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one value.

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For example if I plug in three for n I get three to the fourth which is eighty one.

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So this 81 value is a much more significant value than just this simple plus one right.

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I can have 81 the plus one changes it to 82 but 81 does most of the work here it's much greater than

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this value here of 1.

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So end of fourth is going to have a much greater effect on this whole series than just this plus one.

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So one thing we considered doing is getting rid of this plus one if we do let's talk about here B.

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Sub an R comparison series.

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If we do that what we're left with is 1 over the square root and to the Fourth and this is another clue

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for us.

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This is going to turn out to be a very convenient comparison series when we take the square root of

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end of the fourth we get and squared.

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We get one over and squared.

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Now what we know about this series is that it's a P series that converges because when you use the pre-series

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test you know that if this value here which is P this exponent when P is greater than 1 this series

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will converge.

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So this is a piece with p equals to P is greater than 1 and we know therefore that piece of and converges.

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So B-7 is a convergence sequence.

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Now we can say is that if Subban is less than B-7 then we know that A7 also must converge and you can

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think about it like this if you had a graph like this for example and would just do this really quickly

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these graphs don't represent a 7 or B 7 but if you just think about a convergence series and let's say

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that this series is be Subhan for example and it's converging down here toward the x axis if a submarine

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is always less than the series right it's always underneath it.

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If it's always underneath B-7 then it also must converge.

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It can't divert because then if it diverged at some point it would become greater than be Subhan.

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So if we can show that A-7 is always less than B-7 then we can prove that A7 also converges.

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So let's look at the values of A7 versus B-7.

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If we plug in values here for B 7 and we start with an equals 1 we'll say that when we plug in 1 we

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get 1 squared which is just 1 so the first term is 1.

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When we plug in 2 we get 2 square which is for us we get 1 fourth.

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When we plug in three we get 1 ninth and we kept going you'd see one 16 tier 1 25th etc..

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What we need to do now is compare each term in our series so this was the turn that we got when we plugged

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in 1 to a 7.

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This is the term that we got when we plugged in 1 to be 7.

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This is the term that we got only plugged into to a 7.

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And this is the term that we got when we plugged into to be Subban.

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If we compare these if you plug these square roots into your calculator and you do one with squirted

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to 1 over the square to 17 etc. what we'll find is that the values in B-7 and the B-7 series will always

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be greater than the values in A-7.

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So one is greater than 1 over the square of two.

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One fourth is greater than 1 over the square to 17 when 9 is greater than 1 over the square of 82 et

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cetera.

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So we know that B-7 be Subhan is always greater than a Sabun.

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Therefore we know that a 7 converges since B 7 converges and is always greater than a 7.

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So by the comparison test we've proven that a 7 converges.

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Now we can use the remainder estimate for the integral test to get an estimate of the remainder for

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a 7.

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So we'll call the remainder for a sub n we'll call it our Subhan would be the remainder.

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In this case we're doing 10 terms right.

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The first 10 terms of the series.

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So we're going to call it our subtend by this remainder estimate.

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Theorem we know that our subtend is always going to be less than or equal to the integral from here

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10 because we're doing the first 10 terms to infinity of our comparison series B so so we'll say 1 over

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instead of using end squared.

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We'll just use x squared and then D x.

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Really what this theorem is telling us is that the remainder is going to be less than the value of this

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integral and what it's actually saying is that the remainder of the series B 7 is going to be less than

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this integral.

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However we've already proven that a Sabahan will always be less than be 7.

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So what we have for an inequality here is our subtend which is the remainder of the series.

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A-7 will always be less than or equal to.

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We'll call it T sub and or sub 10 which is going to be the remainder of the comparison series B 7 which

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is always going to be less than or equal to the integral of our comparison series B 7 from this value

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here 10 to infinity.

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So because we plugged in our comparison series B 7 here we know that our remainder for B Subban has

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to be less than or equal to that value that's t subtend.

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And because A7 is always underneath be subtend like we drew in that graph the remainder for Ace habban

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is going to have to be less than the remainder for B-7.

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So we can just go ahead and write this inequality here and then evaluate the integrals so we'll say

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our subtend is less than or equal to when we take the integral.

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Here we can consider this instead of 1 over x squared we'll call it X to the negative 2 we'll move this

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x squared to the numerator.

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Changing the exponent to a negative value then we just add 1 to our exponent negative two plus one is

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negative one.

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So we get X to the negative 1 and then we divide by our new exponent so we divide by negative 1 like

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this and we're going to be evaluating that on the interval 10 to infinity.

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Now of course we can simplify this function.

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Here we have X to the negative 1 in the numerator.

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We can move that to the denominator and change the exponent from a negative to a positive.

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So that just becomes X and the denominator.

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And we have this negative one here in the denominator which we can move to the numerator.

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What we really end up with here is just negative 1 over X that's our integrated value there and then

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we evaluate on this interval.

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So instead of evaluating for intend to infinity the technically correct way to write is we want to say

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that this is less than or equal to the limit as a goes to infinity of negative 1 over X evaluated and

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then the interval.

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Ten to a.

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So we've really just replaced the infinity with a.

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And we're saying we're going to evaluate the limit as a goes to infinity or approaches infinity.

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Of course here when we're taking this definite integral like this we want to plug in our upper limit

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here a first.

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So we're going to say the limit as a goes to infinity we'll plug in a and we'll get negative 1 overlay

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then we'll subtract whatever we get when we plug in our lower limit here 10.

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So we're going to get negative 1 over 10.

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So say negative 1 over 10 like this are negative signs here are going to cancel when we evaluate a at

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infinity and we get this infinite value here in the denominator.

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This whole fraction negative 1 over infinity is going to become 0 because any constant divided by an

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infinitely large value here is always going to tend toward zero.

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So that's going to go away.

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And what you see that we're left with is an estimate for the remainder are sub 10 as being less than

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or equal to one tenth.

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Or we can also write that zero point one either way.

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So that's our final answer for the remainder of the way that we want to write our complete final answer

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is we want to say that some will say a subtend is going to be approximately equal to because remember

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this was an approximation.

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1 point 2 4 8 6 with a remainder value with our sub 10 less than or equal to zero point 1 which we calculated

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using the comparison Series B 7 knowing that be Sabayon also had to be less than 0.1.