Original subtitles

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Too we're going to be talking about how to determine whether or not a series converges absolutely or

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converges conditionally.

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And in this particular video we're going to be doing two different examples trying to determine whether

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or not each of these infinite series converges absolutely or conditionally if at all.

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Now it's important to remember that a series is conditionally convergent if it converges but is not

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absolutely convergent.

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The series is absolutely convergent if the absolute value of the series is convergent And so basically

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what we can what we can remember is that if the absolute value of the series a seven and we would call

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this part of the series here this value this formula for the series.

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This is a subset.

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And this here is a subset.

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Another example of a series is the absolute value of a sub N is convergent.

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If this is convergent then we know that the series is absolutely convergent.

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Otherwise it may not be absolutely convergent.

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This particular part might not be true but this series in general without the absolute value bars may

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be convergent.

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And in that case it would be conditionally convergent.

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So when we're talking about absolute convergence the best way to test for it is using either the ratio

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test or the root test.

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If we can the ratio and root tests for Convergence have absolute convergence built right into the definition

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of those convergence tests.

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So if there's any way that we can use those we want to.

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The ratio test is often a little bit more flexible than the root test in terms of applying it to series

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in general.

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But either one will allow us to determine absolute convergence if the series is set up in a way such

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that we can use the tests.

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So let's go through one example of each.

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So we can start getting a feel for how we can use those tests to determine absolute convergence.

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So in this first example we have the infinite sum from equals 1 to infinity of this series which is

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10 to the power divided by the quantity and plus one at times 4 raised to the power to and plus 1.

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Now this is a perfect candidate for the ratio test.

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And the reason is because the ratio test tells us that the limit L is equal to the limit as an goes

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to infinity of the absolute value of basically a sub and plus 1 divided by a sub n and all that means

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is that we're going to be plugging in and plus 1 everywhere we have N in our original series here and

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we're going to put that in the numerator and these absolute value bars and then we're going to take

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the original function a Subban and put that in the denominator.

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So in other words let's go ahead and substitute and plus 1 everywhere are we have.

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And in our original series so instead of tens the end will get 10 to the end plus 1.

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And then in our denominator here instead of and plus one will substitute and plus 1 here for this end

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and get and plus one plus one or in other words and plus two and then here we'll get four to the two

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times.

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And plus one two times and plus one is two plus two.

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So we'll get to N plus two plus one or in other words two and plus three.

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So we got four race to the two end plus three power.

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Then we divide that by the original series a seven.

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So we divide it by 10 to the N divided by and plus 1 times 4 to the two and plus 1 power.

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Now this is sort of an intermediate step that people skip a lot of times because you're always going

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to get with this test.

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A fraction divided by a fraction and of course when we have that situation we can just take the numerator

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here.

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So ten to the end plus 1 divided by and plus two times 4 to the two and plus three.

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And instead of dividing by this fraction in the denominator we can multiply by it's reciprocal that's

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the same thing.

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So we just flip it upside down and we get.

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And plus 1 times 4 to the two and plus 1 all divided by 10 to the end.

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And we're taking the absolute value of this whole thing.

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Now as you can see knowing that we could have just gone straight to this step and skipped writing out

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this whole thing here.

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So either way however you want to do it is is fine.

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So in this case now that we have the fractions written this way we're just looking to match up values

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from the numerator and denominator and what that means is basically we're looking for like bases or

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terms that are similar to one another.

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So for example we have 10 to the end plus one here and we have 10 to the end.

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We've got the same base of 10 just a different exponent.

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So remember that when we have fractions like this we're going to kind of pair these up together when

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we have for example let's just take an easy one right.

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Tend to the third over 10 squared we're going to simplify this by subtracting the exponent in the denominator

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from the exponent in the numerator.

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So three minus two is one this becomes tend to the first power is the simplified version of tend to

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the third over 10 square.

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Same thing here we have tend to the end plus one over tend to the N.

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So the way that we simplify that tend to the end plus one over tend to the end.

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We subtract and from and plus once we get and plus one minus N and in that case and plus one minus N

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we get the ends to cancel.

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And we're just left with one.

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So this is 10 to the first power.

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So you can see how that.

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So what we're going to be left with for that particular term is just ten to the first power in the numerator

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so 10 of the first power there.

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Now let's go ahead and match up 4 to the two and plus one in the numerator and four to the two and plus

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three in the denominator.

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If we take two and plus one from the numerator and we subtract what's in the denominator 2 and plus

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3 what we'll get is to N plus one minus two and minus three are two ends cancel and we're left with

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1 minus three which is a negative 2.

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So what that tells us because this value is negative is that we're just left with four race to the negative

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two.

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Or in other words one over four to the positive to move that to the denominator.

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We get a positive value.

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We're just going we left with four squared in the denominator.

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So we put this in the denominator for squared like that.

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That's all that's left of this term.

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And then of course we have that multiplied by Plus 1 over and plus 2 which we can't simplify there's

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no exponents there to simplify.

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Those are just terms that are remaining and to evaluate this limit we can go ahead and pull out the

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10 over four squared remember that four squared is 16.

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So we essentially have 10 over 16 here or in other words 5 over 8.

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When we reduce it.

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So what we're left with is the limit is equal to five eighths times the limit as and goes to infinity

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of the absolute value of plus 1 over and plus 2.

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And when we have just a rational function like this polynomials in the numerator and denominator the

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easiest way to evaluate this infinite limit is to divide through both the numerator and denominator

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by the highest degree and Terman in this case that's And to the first power.

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So we want to multiply by 1 over into the first divided by 1 over to the first like this.

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And what we're left with then is our equals five eighths times the limit as and goes to infinity of.

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And at times 1 over end just gives us one.

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So we have one plus 1 times 1 over and gives us 1 over and.

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And then in the denominator end times 1 over and gives us 1 2 times 1 over and gives us 2 over and.

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And the reason that we do this is because now we have these values of n in the denominator.

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When we have a constant like one or two here divided by an end is going to infinity.

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This denominator becomes extremely large and these two terms here are eventually going to tend toward

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zero and becomes larger and larger and larger.

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So these two are both going to go away and become 0 and as you can see all that we're left with is just

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1 over 1 or 1 the limit as and goes to infinity of 1 is just 1 itself.

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So our limit is equal to five eighths times one or just five eighths.

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Now the ratio test tells us here's where the conclusion of the ratio test comes in the ratio test tells

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us that whatever result we have here if it is less than 1 then a series is absolutely convergent.

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If the value is equal to 1 the test is inconclusive the value is greater than 1.

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Then we know that the series diverges.

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But in this case the value is less than 1 5 8 is less than 1.

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So by the ratio test this series here is absolutely convergent.

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It converges Absolutely.

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So we looked at an example of how to use the ratio test to determine whether or not a series converges.

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Absolutely.

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Now let's take a look an example of how to use the root test to determine whether or not the series

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converges.

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Absolutely.

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We're going to look at this infinite sum here from unequals went to infinity of the quantity and squared

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plus one divided by two n squared plus one all raised to the end power.

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Now this is a perfect candidate for the root test because the root test tells us that the limit is equal

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to the limit as an goes to infinity of the end through of the absolute value of the series a 7.

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So in other words remember that our series A seben is represented by the function here.

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So we're going to put that inside absolute value bars and we're going to take the nth root of it.

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Keep in mind that when we take the and through this and through here it's just going to cancel out this

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and through right here and that's what makes this particular series a perfect candidate for the root

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test.

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When ever you can raise everything in the entire series whenever you can put the entire series in parentheses

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like this and raise it to the end power or the two and power or something like that.

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Then the root test is a great test to use because you can get this xponent here to cancel thereby significantly

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reducing the complexity of the series.

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So given that we're going to say that the limit is equal to the limit as and goes to infinity of the

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absolute value of our series so we have squared plus one divided by two and squared plus one.

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And keep in mind that the series here we had it raised to the power like this raised to the power well

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this and throughout the series is basically raising this whole thing to the one over and that's why

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they cancel.

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Because and times one over end is just one.

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So this goes away and we're just left with the absolute value of an squared plus one over two and squared

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plus one.

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Given that we're going to do the same kind of thing that we did with our ratio test where we multiply

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both numerator and denominator by the highest degree and variable in this case that end squared right.

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And to the power of two is the highest exponent on any and variable in this sequence here this series.

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So we're going to multiply by one over and squared divided by one over and squared.

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And what that's going to give us limit equals l equals limit as and goes to infinity.

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What that's going to give us and squared at times 1 over and square we get the squares to cancel and

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we're left with one plus one times one over and squared is one over and squared and then in the denominator

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same thing we get the squares here to cancel and we're left with two plus one times one over and squared

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is just won over and squared.

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And now we have that same situation where as and becomes very very very large goes towards infinity.

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These two small fractions here will become 0 1 if you on your calculator take one and you divide it

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by a very large number like one million or 10 million.

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Your calculator will actually give you an answer of zero because this number is going to become so incredibly

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small that it becomes insignificant and we can just call it zero.

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They both cancel as you can see obviously we're just left with L equals the limit as and goes to infinity

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of the absolute value of one half.

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Well the is of 1 1/2 is just one half.

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The limit is and goes to infinity of 1 half is still just one half.

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There is no value we have to plug in for more.

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So what that tells us is that our limit L is equal to one half.

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And similarly with the root test as with the ratio test when with the root test L is less than 1.

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We know that the series converges Absolutely.

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So we can call this series absolutely convergent.

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We'll label this up here also.

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Absolutely convergent.

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And keep in mind that it's going to be the same thing if we were to get out.

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Equals 1.

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The test would be inconclusive if we get a greater than 1 then we know by the root test that the series

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diverges.

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But because the value is the value of the limit L is less than 1.

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We know that by the root test the series is absolutely convergent.