Uploaded by TheIntegralCALC on 31.10.2011

Transcript:

Hi, everyone. Welcome back to integralcalc.com. Today we’re going to be talking about using

the ratio test to determine whether or not a series converges or diverges. So I’ve

taken this screen-grab from a table that I have in my website that talks about the different

convergence tests and what it shows us is that we have a series, we’ll call it a sub

n. In order to determine whether or not it converges, what we can do using the ratio

test is we can plug k sub 1, in this case our variable is k, in for k and take that

entire function and divide it by our original function. What we get when we simplify and

take the limit as the variable goes to infinity, as k goes to infinity. The answer that we’re

going to get if it is less than one, then the series converges and if it’s greater

than one, then the series diverges. So let’s take a look at what this looks

like. So in this case, we have the sum of 3 to the k divided by k squared. from k equals

1 to infinity. So what we want to do like we mentioned before is plug in k sub 1 every

time we see the variable k. So to use the ratio test, we’re going to say the limit

as k goes to infinity of 3 to the k plus 1. Again we’re placing k plus 1 everywhere

we see the variable k divided by k plus 1 squared. We’re going to divide this entire

function by the original function. So 3k divided by k squared. And now it’s just a matter

of simplifying this and then finding the limit as k goes to infinity. So in order to simplify,

the first thing we’ll do is turn this into a multiplication problem instead of a division

problem. Remember that when you divide by a fraction, it’s the same thing as multiplying

by its inverse. So what we can say is 3 to the k plus 1 divided by k plus 1 squared.

And instead of dividing by this fraction, we’ll multiply by the fraction’s inverse.

It’s basically flipping it upside down. So instead of k squared being in the denominator,

we put it in the numerator and instead of 3 to the k being in the numerator, we’re

bringing it to the denominator and now we’re multiplying instead of dividing.

So we can reorganize these terms, somewhat. We kind of pair them up, similar terms together

so we get k plus 1 over 3k times k squared over k plus 1 squared. So the reason I did

this is because I want to go ahead and simplify the first part here, 3 to the k plus 1 divided

by 3 to the k. Remember that when you have something in the numerator that has the same

base as another term in the denominator, in our case the bases are both equal to 3, you

can subtract the exponent in the denominator from the exponent in the numerator. This is

no different than saying x to the 4 over x cubed. We all know that that is equal to x.

The reason is because we’re taking 4 minus 3 and we get 1. This is x to the first. So

we subtract the exponent in the numerator from the exponent in the denominator. In this

case, we’ve got k plus 1, the exponent in the numerator and then we subtract the exponent

in the denominator. What we can see there is that we can get the k’s to cancel and

we’re just going to be left with 1. So we’ll have 3 to the first power. 3 to the first

power is of course just 3 so we can now pull 3 out in front of these limit here. So now,

we’re just taking 3 times the limit as k goes to infinity of k squared over k plus

1 squared. And what we want to do now, I’m going to

go ahead and multiply out the denominator so that we can see what we’ve got a little

bit more clearly. So the limit as k goes to infinity of k squared over and when we multiply

out the denominator, which is k plus 1 times k plus 1, we’ll get k squared plus 2k plus

1. And now, this is something that we do really commonly with these types of sequences and

series problems. In order to evaluate the limit as k goes to infinity, we’ll go ahead

and divide through the entire function here by k squared, which is the highest degree

term. k squared is the term with the largest exponent so if we had a k cubed somewhere

in here, we’ll want to divide by k cubed. The reason we do the largest one is because

we don’t want k to remain in the numerator of anyone of these individual terms. And remember

we’ll be dividing every single term by k squared. So when we divide k squared by k

squared, we’ll get 1. When we divide k squared in the denominator by k squared, we’ll get

1. 2k divided by k squared is going to give us 2/k. and 1 divided by k squared gives us

1 over k squared. So now, we can evaluate because when we plug

infinity in for k, this term here, 2/k will go to zero because the denominator of that

particular fraction becomes very, very, very large. And when the denominator goes to infinity

and becomes large, the entire term is going to approach zero so that will become zero.

This term here, 1 over k squared again the denominator will become infinitely large and

that term will approach zero. So what we’re left with is 3 times the limit as k approaches

infinity of 1 over 1, which is just 1. So our answer here is 3. If we return back to

our original formula or the ratio test that we’re using, we can see here that the function

converges if L is less than one and diverges if L is greater than 1. We’ve just found

that L is equal to 3. So since 3 is larger than 1, that means that the series is going

to diverge. So we’ll go ahead and say, diverges by the ratio test. Because we found the L

is equal to 3 and 3 is greater than 1 which is what the ratio test asks us to apply.

So that’s it. I hope that video helped you guys and I will see you in the next one. Bye!

the ratio test to determine whether or not a series converges or diverges. So I’ve

taken this screen-grab from a table that I have in my website that talks about the different

convergence tests and what it shows us is that we have a series, we’ll call it a sub

n. In order to determine whether or not it converges, what we can do using the ratio

test is we can plug k sub 1, in this case our variable is k, in for k and take that

entire function and divide it by our original function. What we get when we simplify and

take the limit as the variable goes to infinity, as k goes to infinity. The answer that we’re

going to get if it is less than one, then the series converges and if it’s greater

than one, then the series diverges. So let’s take a look at what this looks

like. So in this case, we have the sum of 3 to the k divided by k squared. from k equals

1 to infinity. So what we want to do like we mentioned before is plug in k sub 1 every

time we see the variable k. So to use the ratio test, we’re going to say the limit

as k goes to infinity of 3 to the k plus 1. Again we’re placing k plus 1 everywhere

we see the variable k divided by k plus 1 squared. We’re going to divide this entire

function by the original function. So 3k divided by k squared. And now it’s just a matter

of simplifying this and then finding the limit as k goes to infinity. So in order to simplify,

the first thing we’ll do is turn this into a multiplication problem instead of a division

problem. Remember that when you divide by a fraction, it’s the same thing as multiplying

by its inverse. So what we can say is 3 to the k plus 1 divided by k plus 1 squared.

And instead of dividing by this fraction, we’ll multiply by the fraction’s inverse.

It’s basically flipping it upside down. So instead of k squared being in the denominator,

we put it in the numerator and instead of 3 to the k being in the numerator, we’re

bringing it to the denominator and now we’re multiplying instead of dividing.

So we can reorganize these terms, somewhat. We kind of pair them up, similar terms together

so we get k plus 1 over 3k times k squared over k plus 1 squared. So the reason I did

this is because I want to go ahead and simplify the first part here, 3 to the k plus 1 divided

by 3 to the k. Remember that when you have something in the numerator that has the same

base as another term in the denominator, in our case the bases are both equal to 3, you

can subtract the exponent in the denominator from the exponent in the numerator. This is

no different than saying x to the 4 over x cubed. We all know that that is equal to x.

The reason is because we’re taking 4 minus 3 and we get 1. This is x to the first. So

we subtract the exponent in the numerator from the exponent in the denominator. In this

case, we’ve got k plus 1, the exponent in the numerator and then we subtract the exponent

in the denominator. What we can see there is that we can get the k’s to cancel and

we’re just going to be left with 1. So we’ll have 3 to the first power. 3 to the first

power is of course just 3 so we can now pull 3 out in front of these limit here. So now,

we’re just taking 3 times the limit as k goes to infinity of k squared over k plus

1 squared. And what we want to do now, I’m going to

go ahead and multiply out the denominator so that we can see what we’ve got a little

bit more clearly. So the limit as k goes to infinity of k squared over and when we multiply

out the denominator, which is k plus 1 times k plus 1, we’ll get k squared plus 2k plus

1. And now, this is something that we do really commonly with these types of sequences and

series problems. In order to evaluate the limit as k goes to infinity, we’ll go ahead

and divide through the entire function here by k squared, which is the highest degree

term. k squared is the term with the largest exponent so if we had a k cubed somewhere

in here, we’ll want to divide by k cubed. The reason we do the largest one is because

we don’t want k to remain in the numerator of anyone of these individual terms. And remember

we’ll be dividing every single term by k squared. So when we divide k squared by k

squared, we’ll get 1. When we divide k squared in the denominator by k squared, we’ll get

1. 2k divided by k squared is going to give us 2/k. and 1 divided by k squared gives us

1 over k squared. So now, we can evaluate because when we plug

infinity in for k, this term here, 2/k will go to zero because the denominator of that

particular fraction becomes very, very, very large. And when the denominator goes to infinity

and becomes large, the entire term is going to approach zero so that will become zero.

This term here, 1 over k squared again the denominator will become infinitely large and

that term will approach zero. So what we’re left with is 3 times the limit as k approaches

infinity of 1 over 1, which is just 1. So our answer here is 3. If we return back to

our original formula or the ratio test that we’re using, we can see here that the function

converges if L is less than one and diverges if L is greater than 1. We’ve just found

that L is equal to 3. So since 3 is larger than 1, that means that the series is going

to diverge. So we’ll go ahead and say, diverges by the ratio test. Because we found the L

is equal to 3 and 3 is greater than 1 which is what the ratio test asks us to apply.

So that’s it. I hope that video helped you guys and I will see you in the next one. Bye!