All language subtitles for 309 Ratio test with factorials-subtitle-en

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

converges.

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

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

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

of a 7 plus one divided by a Sabun.

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

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

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

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

value of N with end plus 1 instead.

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

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

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

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

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

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

So again everything hinges on this value.

So that's the value we need to find.

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

directly from this formula here.

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

So the absolute value of a seven plus one.

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

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

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

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

plus 1.

So this value right here is a seven plus one.

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

two to the N divided by an factorial.

We're taking the absolute value of this quotient.

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

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

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

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

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

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

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

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

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

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

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

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

And we have to to the end.

The base is two.

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

Here we have two factorial terms so those are similar.

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

the value because these are multiplied together.

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

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

and plus 1 factorial.

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

they both raise two different exponent.

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

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

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

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

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

So this whole fraction here becomes two.

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

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

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

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

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

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

one right.

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

for three is the same thing as five minus two.

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

We have five minus four.

So really we're just starting with five.

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

together.

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

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

started here with five.

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

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

like this dot dot dot.

And it would go on for ever.

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

5 and my 6 and 7 on into infinity.

So that's what and factorial represents.

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

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

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

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

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

1.

Then we get 5 minus 2.

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

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

So Id say and minus one.

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

that.

And we had the absolute value here.

So remember in the numerator we started with factorial.

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

we started with and plus 1.

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

on into infinity.

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

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

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

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

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

cancel.

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

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

left with one.

And then in the denominator we have.

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

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

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

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

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

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

And to the first.

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

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

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

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

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

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

absolute value.

So we just multiply across numerator and denominator like this.

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

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

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

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

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

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

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

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

So this is going to go to zero.

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

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

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

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

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

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

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

The only special value of L is equal to 1.

One is the value that everything hinges on.

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

since zero is less than 1.

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

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