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In this video we're talking about how to use the alternating series tests to say whether or not a series

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converges.

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And before we go any further it's really important to remember that the alternating series test can

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never prove divergence.

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So if you're going to use the alternating series Test the only conclusions you're going to be able to

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draw are that the series converges or that the test is inconclusive and you're going to have to try

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to use a different test to determine convergence or divergence.

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So with the alternating series Test it's convergence or nothing.

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So in this particular problem we've been given the series given by some from equals one to infinity

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of negative one to the end plus one multiplied by squared of and plus one minus square root of n.

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So this negative 1 to the N plus 1 term tells us that this is an alternating series of negative 1 to

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the end plus 1 and minus 1 or even just to the end.

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This term is going to make the entire series alternate sign.

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So it's got alternate positive negative positive negative positive negative.

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So it's an alternating series.

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The important thing to know about an all training series and for the alternating series tests is that

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really most of the time we're just focused on the value outside of this negative one term.

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So we're really going to be looking at is this value right here.

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And when it comes to the alternating series tests we say for an altering series a seven and a sub and

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the series a seven is going to be this value we've put brackets around so everything but this negative

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one term that makes the series alternate.

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So we drop everything but this negative one term for A7 and what we say is that if the series is decreasing

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that's what this inequality means that a sub and plus one is less than a seben and they're both greater

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than zero.

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So both the series seven plus one and a seven are positive but the Southern plus one term is less than

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the seven term.

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So the series is decreasing.

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So if we can show that and we can show that the infinite limit of a sedan is equal to zero then by the

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alternating series Test we can prove that a converges.

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So we're really just tackling these two conditions here.

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So let's go ahead and start with this inequality that says A7 plus one has to be less than a seven.

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There's a couple of ways to do this and I'll show you both of them.

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The first one is just to expand the first few terms of the series and compare them across both.

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So if we go ahead and say that a seben is going to be equal to the square root of and plus 1 minus the

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square root of and we're just taking that directly from our original series here so that's a seven then

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a sub and plus 1 so a then plus 1.

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What we want to do is replace every value with an end plus 1 value.

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So instead of n right here we put an end plus 1.

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We keep the other plus 1.

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So we've got that there and then we just say minus the square root of and plus 1 because we replace

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n with end plus 1.

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So what this simplifies to then is the square root of and plus 2 minus the square root of and plus 1.

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So this is the series a seven plus one.

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This is the series a Sabrine.

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So if we start plugging in some values.

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Notice that the index of the series starts at 1 so end starts at 1.

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So if we start plugging in values for an equals 1 and equals 2 3 4 etc. here's or we're going to get.

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So plugging in one we get one plus one is two.

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So we get the square root of two minus the square root of 1 because we plug one in here then we're going

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to add to that whatever we get when we plug in N equals 2.

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So here we'll get the square root of 3 minus the square root of 2.

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Then we're going to add to that plugging in here three we're going to get the square root of 4.

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Minus the square root of 3 and we could keep going.

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Let's look at that same process for a seven plus one.

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So here plugging in N equals 1.

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We're going to get the square root of 3 minus the square root of 2.

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Then plugging in equals two.

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Get the square root of four minus the square root of three.

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And then pugging in any three will get the square root of five minus the square root of four.

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And both of these would continue on indefinitely as we plugged in N equals 4 5 6 7 et cetera.

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But what we see happening here is interesting because if we look at this a 7 series we notice that we

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have a positive root too.

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And then later on in the end equals to term a negative root to.

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Here we've got positive root 3.

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And then here in the end equals three term a negative root 3.

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And that pattern would continue over in the end equals four term.

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We would end up with a negative root for term.

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So here we're going to get these values to cancel.

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We're going to get these values to cancel the square root of four term would cancel with a negative

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square root of 4.

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Later on down the road the only thing that's going to remain then is this negative square root of 1.

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We know that the square root of 1 is one.

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So the result of this it's just a negative one here if we look at a seven plus one what we see is a

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similar pattern so we're going to get squared of three cancell with negative fruit 3.

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We're going to get squared of four to cancel with negative fruit for the square to five would cancel

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with a term later on down the line.

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So all we're left with then is this negative square root of two value.

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So what we could say is that the series is going to be equal to negative route to well negative one

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is greater than negative square root 2.

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So that shows that a seben is greater than a seven plus one.

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You're not always going to get the series to work out that cleanly though.

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So the safest way to make sure that the series is decreasing is to actually take the derivative of the

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original series.

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Remember that the original series is this squared of N plus 1 minus square root of n.

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So if we say Y is equal to squared of N plus 1 minus squared even then if we wanted to take the derivative

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we would get y prime or d y over d n is going to be equal to and then taking the derivative of this

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series.

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Remember that the square root of plus 1 is the same thing as quantity and plus 1 race to the one half

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power.

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So if we want to take the derivative we just use power rule.

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We bring that one half ex-porn him down in front and we get 1 1/2 times and plus 1 and then subtracting

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1 from the exponent we get a negative 1 1/2.

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Same thing year the squirt of end is the same as and to the one half power.

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So we bring that one half down in front and we get minus one half.

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And to the negative 1 1/2.

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Now if we simplify this what we end up with for this first term is 1 divided by two times the square

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root of plus 1 because this negative one half becomes a positive one half when we move this and plus

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one to the denominator.

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So we have and plus one to the positive one half in the denominator.

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But of course and plus one to the positive one half is the same as squared of plus 1 which is how we

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get that square root in the denominator and then same thing here we would end up with a minus 1 over

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two times the square root of N.

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Now in order to find a common denominator here we would want to multiply this first fraction by the

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squared of an over the square of n we would want to multiply the second fraction by the square root

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of plus 1 over the square root of and plus 1.

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Because when we do that here in the numerator we'll get square root of N then we'll have this minus

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one times square and plus 1 or just minus square root of and plus 1 in the denominator.

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Now we have a common denominator of two times the squared of and times the square root of and plus 1.

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If we then multiply by the conjugate of the numerator.

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Remember that the conjugate of the numerator is the same two terms but with the sign in the middle changed

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from a negative to a positive in this case.

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So squared of N plus square root of and plus 1 instead of minus.

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And then the same thing in the denominator because if the numerator and denominator were different we

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wouldn't be multiplying by 1.

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So we would be changing the value but here because the numerator and denominator are the same.

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This is just like multiplying by 1 so we're not actually changing the value.

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But what happens then when we multiply by the conjugate is we get squared event times square to then

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which is.

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And here we get squared even time squared of N plus 1.

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But then we have minus the same thing squared of end time squared event plus 1.

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So those middle terms cancel and we're just left with minus the squared of and plus 1 times the squared

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of and plus 1 or minus and plus 1.

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And then in our denominator if we multiply this two and times route and plus 1 times root n we'll get

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the root and to cancel with the root and giving us just.

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And so we have two N times the squared of N plus 1.

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Then we have a plus to square root n times and plus 1.

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If we distribute this negative sign in the numerator what we end up with is and minus and minus 1.

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So let's go ahead and change this whole thing to and minus and minus 1.

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That will allow us to cancel our ends we have and minus.

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And so what we're left with then finally is just a negative one divided by two and square root of end

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plus 1 plus two times and plus one Times Square root of n.

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So now here's what we see.

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We know that the index for an starts at 1.

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Which means here that the lowest value the end can possibly have is 1.

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It can have values of 1 2 3 4 all the way on up to infinity but it's always going to be a positive number

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which means that 2 times n is always going to be a positive number and plus one is always going to be

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a positive number and taking the square root of a positive number will always give us a positive number

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and plus one will always be a positive number.

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And when we multiply that by 2 we'll still always get a positive.

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And the square root of N here will always be a positive number.

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So we have a positive times a positive plus a positive times a positive.

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So it's safe to say that the denominator of this fraction will always be positive.

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On the other hand the numerator is a constant negative one.

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So the numerator is always going to be a negative number.

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So what we can say then is that the derivative D-y dn the derivative is going to be a negative divided

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by a positive.

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The result of which will always be a negative.

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So the derivative is always negative.

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And remember that the derivative represents the slope.

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So if the derivative is always negative the slope is always negative which means the series is always

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decreasing and if the series is always decreasing then a seven plus one is always going to be less than

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a 7.

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So based on that we can go ahead and check off these conditions.

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We know that both series are going to be greater than zero.

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They're always going to be positive but we also know that a seven plus one is always going to be less

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than a 7 because the series is decreasing.

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Our only condition left to check then is that the limit is and goes to infinity of the original series

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is equal to zero.

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So we just want to take the limit as and approaches infinity of our original series.

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Square root and plus one minus the square root of N..

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If this limit is 0 then because we've already met this first condition will have met the second condition.

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Also we'll be able to prove that a 7 converges well just by looking at this we can see that the limit

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is going to tend towards zero because infinity plus one is not significantly different than infinity

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which means the square root of infinity plus one is not going to be significantly different than the

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square of infinity.

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Basically we're just taking the square of infinity minus the square of infinity which would be zero.

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So the limit will be zero.

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If you want to check that one thing you could do is multiply by the context here so we'd say the square

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root of and plus 1 instead of minus we do plus the square root of n.

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And then of course the denominator would be the same thing plus the squared of n.

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And if we did that then we would multiply out these two terms by these two terms in the numerator here

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so square root and plus 1 Time Square and plus 1 gives us and plus 1 we'd have squared and plus one

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time squared n but then minus squared and Times Square and plus 1.

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Those two terms that are on the net to zero.

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So then we just have minus route and times route n or minus N and then in the denominator we would end

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up with root and plus 1 plus root.

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And of course the limit as and approaches infinity.

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So then here you can see we're going to get an minus and to go to zero.

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So what we're left with then is just the limit as and goes to infinity of 1 divided by square root and

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plus 1 plus square root n.

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And here we can clearly see that if we were to evaluate as and goes to infinity we get infinity plus

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one or an extremely large number here taking the square root would stop a very large number plus the

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square root of infinity.

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Again a very large number.

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So what we basically end up with is 1 divided by infinity which is of course zero.

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So we can go ahead and say that this condition has been met as well since we've met all three conditions.

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We can say that by the alternating series Test this original series that we were given is going to converge

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and that's how you use the alternating series test to determine the convergence of an alternating series.