Ratio Test - Calculus


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!