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In this video we're going to be talking about how to use the ratio test to say whether or not a series

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converges.

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And in this example we have a series that has effect Tauriel involved so we have factorial in the denominator

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of our series the ratio test is just one of our convergence tests and it all hinges on this value here

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of L and L is defined as the limit is and goes to infinity of the absolute value the apps I use important

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of a 7 plus one divided by a Sabun.

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Well the first thing to know is that there's a sub N value in the denominator.

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Right here is just the series a sub and the original series.

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So we'll plug in this series here for a sub and just plug this value directly in.

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For A-7 a seven plus one is the value that we get when we take this original series and we replace every

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value of N with end plus 1 instead.

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So whatever value we get when we do that we plug that in for a seven plus one and then we have this

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quotient of series here and we take the EPS value and then we take the limit whatever we get.

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Over here is going to be the value of L and the ratio test tells us that if is less than 1 in the series

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converges if Al is greater than 1 then the series diverges.

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And if L is equal to one then the ratio test in particular is inconclusive and we have to use another

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test or we might just have to say that we can't determine convergence.

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So again everything hinges on this value.

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So that's the value we need to find.

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So we're going to go ahead and say Al is equal to the limit as and goes to infinity you are taking this

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directly from this formula here.

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And then we're going to say the absolute value of a seven plus one.

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So the absolute value of a seven plus one.

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So a seven plus one is again what we're going to get when we plug and plus one in for n in the original

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series so instead of 2 to the end we'll get to the end plus one.

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And then instead of end factorial we're going to get an plus 1 factorial.

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And it's important to put these parentheses here to indicate that we're taking the factorial of and

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plus 1.

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So this value right here is a seven plus one.

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And then we're going to divide that by a submarine which again is just the original series so that's

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two to the N divided by an factorial.

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We're taking the absolute value of this quotient.

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Now a lot of people like to skip this step and go straight to the second step that we're writing out

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here because remember we just have a fraction divided by a fraction.

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And when that's the case where we can do is we take the fraction in the numerator we leave that as is

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we have to to the end plus 1 divided by quantity and plus one factorial.

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And then instead of dividing by this fraction we can multiply by it's reciprocal that's the same thing.

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So we flip this upside down instead of two to the end divided by and vectorial we get and factorial

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divided by 2 to the end and we haven't changed the value of this quotient at all.

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So now we have this here and you can go straight to the step skipping this one if you want to.

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So instead of division do multiplication and then just do the reciprocal of the n.

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But either way this is where you want to get to.

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And then at this point what we want to do is pair together similar terms and usually that's going to

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mean similar basis so what we can say is that we have to add to the plus one the has to.

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And we have to to the end.

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The base is two.

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So because these have the same base we want a pair of them together.

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Here we have two factorial terms so those are similar.

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We want to pair those together and we're going to rewrite this as the limit as and goes to infinity

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the value because these are multiplied together.

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We can just swap the denominators and it doesn't change the value at all.

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So we end up with two to the end plus one over two to the n multiplied by and factorial divided by quantity

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and plus 1 factorial.

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So now the reason this is helpful is because when we have the same base in the numerator and denominator

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they both raise two different exponent.

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What we can do is subtract the exponent in the denominator from the exponent in the numerator.

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So this value here becomes two to the end plus one the exponent in the numerator.

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Then we just subtract the exponent from the denominator so we end up with and plus 1 minus.

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And for the new exponent well and minus and those cancel that zero.

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So we just have two to the first power which is just 2.

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So this whole fraction here becomes two.

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And because the limit as end goes to infinity doesn't affect the value.

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There's no value involved anymore it's just two we can pull that too out in front of the limit so we

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can say L is equal to two times the limit as and goes to infinity of the absolute value.

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And now here we have an factorial over and plus 1 factorial.

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Now in order to simplify this remaining quotient we have to remember is what a factorial really means.

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So for example if we have five factorial that's equal to five times four times three times two times

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one right.

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So we have this fact you hear five we can call for five minus one right because five minutes one is

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for three is the same thing as five minus two.

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Two is the same thing as five minus three and one is the same thing as five miles for us.

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We have five minus four.

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So really we're just starting with five.

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And then we subtract one we subtract two we subtract three we subtract four we multiply those things

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together.

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So when we have an factorial Let's go ahead and say two times the limit.

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And it goes to infinity when we have N factorial we just start with an in the same way that we just

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started here with five.

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So we start with N and then we multiply that by and minus 1 because here we started with 5 only 5 minus

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1 so we're going to do and minus one and we're going to do and minus two and minus three and minus four

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like this dot dot dot.

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And it would go on for ever.

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We don't know the value of N so we just write dot dot dot because that would continue forever and minus

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5 and my 6 and 7 on into infinity.

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So that's what and factorial represents.

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Now we want to figure out what and plus one factorial represents.

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Well obviously we're just going to start with the original value and plus 1 so we have and plus one

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and then we're going to subtract one just like we do here we did five minus one.

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So and plus one minus one right and plus one minus one the plus one and the minus one cancel and we're

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just left with n so we multiply by n then we subtracted two right in our example here with 5 5 5 minus

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1.

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Then we get 5 minus 2.

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So we would have and plus one and then this time minus 2.

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Well here one minus 2 is in negative ones we end up with and minus 1.

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So Id say and minus one.

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And if we kept going we would get again and minus two and minus three on forever and minus for that

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that.

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And we had the absolute value here.

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So remember in the numerator we started with factorial.

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So here we started with and we said and and then and minus one minus two my three most Mosport here

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we started with and plus 1.

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So we start with and plus 1 and then we subtract one each time and we end up with and minus 1 and 2

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on into infinity.

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So we write those terms out and then what we see is that we can make a lot of cancellations.

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So in this particular problem we have an and an N given and minus one in and minus one.

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These two are going to cancel these who are going to cancel these two are going to cancel.

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And what we see is that if we kept going we would always be able to continue cancelling terms we'd have

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and minus 5 and both numerator and denominator and minus 6 and 7 and minus 8 and we always be able to

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cancel.

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So the only thing we're left with is this end plus one in the denominator so we have two times the limit

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as and goes to infinity and the absolute value since everything canceled in the numerator we're just

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left with one.

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And then in the denominator we have.

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And plus one that's the only thing that's remaining.

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So that's how you simplify factorials when using the ratio test.

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You always do it the same way you write out these factorials to figure out which terms you can cancel

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and then that'll simplify and get rid of the factorial so you can actually evaluate the limit.

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Now at this point when I want to do is divide every term in the numerator and denominator by the largest

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degree variable in the denominator so the largest degree very common denominator is just this and variable.

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And to the first.

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So what I want to do is divide every term in both the numerator and denominator by end of the first.

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In other words I want to multiply by one over and I want to multiply by 1 over n like this.

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Really we haven't changed the value because one over end divided by one over and it's just one when

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we have the same value in the numerator and denominator this fractions equal to 1.

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So really we're just multiplying by 1 so it doesn't change the value.

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But this is convenient because what we're going to get here is two times the limit as and goes to infinity

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absolute value.

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So we just multiply across numerator and denominator like this.

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So we say 1 times 1 over and is 1 over an in the denominator here we have to distribute this one over

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n across both terms so n times 1 over and gets the ends to cancel and we're just left with one and then

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one turns one over and his one over end we have one over n like this.

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And now when we evaluate the limit as and goes to infinity remember that when we have a constant divided

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by infinity for example this fraction here in the numerator 1 divided by n that be 1 divided by infinity

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we'd have a constant of 1 in the numerator and infinity in the denominator.

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That's going to go toward zero because the denominator becomes very very large.

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So the fraction itself becomes very very small and the limit of that is zero.

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So this is going to go to zero.

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And this is going to go to zero when we take the limit.

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So what we're left with then is L as equal to two we're evaluating the limit so we can get rid of this

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limit notation and we have absolute value of zero divided by 1 plus zero.

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Well obviously since we have zero in the numerator this whole fraction is going to go to zero.

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The absolute value of zero we can still call zero sum zero times to is zero and we have l equal to zero.

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Now don't be thrown off of ELAS equal to zero.

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That's not a special value of L when it comes to the ratio test.

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The only special value of L is equal to 1.

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One is the value that everything hinges on.

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So the fact that 0 is less than 1 right if else less than 1 then by the ratio test the series converges

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since zero is less than 1.

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We can say that in this particular case Al is less than 1 and therefore that this particular series

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this two to the end over and factorial is a convergent series by the ratio test.