All right, in this video I'm going to prove that the sequence given here is divergent. There's more
than one way to prove that a sequence diverges. One way is to use the definition of how a sequence
converges and arrive eventually at a contradiction whereby you can conclude that because the
definition of convergence leads to or results in a contradiction, the sequence therefore is
divergent. That's proof by contradiction. A more direct way that I'm going to use in this video is
to just use one of the definitions of divergence for a sequence and that is that if you have two
or more subsequences that converge to different values. Say for example, the odd numbers converge
to -1 whereas the even numbers converge to 1, then that sequence by definition is divergent. Let's go
ahead and get started with that. We're going to write proof. We will prove that the sequence Xn
is equal to -1^n times n minus 1 over 4 n plus 3 diverges by showing two subsequences converge to
different values. If we're going to prove that two subsequences converge to different values,
you kind of have to think what subsequences you want to select in your proof. Anytime you have
-1 to the n you know that for even values that's going to be 1 and for odd values that value is
-1. That's a good place to start thinking about subsequences. And yeah in this case, it does work
that if you just look at the even subsequences and the odd, they are different. Sometimes it
won't necessarily be odd or even, you might have to do a different type of pair of subsequences
but anyway let's do the odd and even examples. For all k in the naturals, then X of 2k--that would be
all even values--is going to be...and then you just plug in 2k into your n for your sequence.
That's going to be -1 to the 2k times 2k minus 1 over 4 times 2k plus 3. And then you're going to
do 2k plus 1. It could be 2k minus 1 but I usually just do plus 1, the next value over. That's going
to represent any odd value. Actually this would start--because the naturals start at 1--this would
be odd values starting at 3 and up but it doesn't really matter because you're just going to show
that two sequences, two subsequences, diverged. It doesn't necessarily have to be all the odd values.
You don't necessarily have to start at 1. All right, but let's just plug in 2k plus 1. That's
going to be -1 to the 2k plus 1 times 2k plus 1 minus 1 over 4 times 2k plus 1 plus 3. Another
thing that you have to think about then is how are you going to prove that these subsequences
converge. Everything looks a little bit--well it's not simplified yet. What you're going to just do
is simplify and rearrange the terms as much as you can to be able to take a limit and show that
they do converge to a real value. For that, we're going to simplify that as...This is always going
to be 1. That just simplifies as 1 times 2k minus 1 over 8k plus 3. We still can't simplify because
we have k in the numerator and the denominator. What we can do is multiply by 1 over k in the
top and bottom and that gives us 2 minus 1 over k over 8 plus 3 over k. If you think about it,
if you take the limit, 1 over a variable that goes to infinity is going to be 0 and so it's the same
thing for the numerator and the denominator. This will converge to a real value. We don't have to
worry about what it is yet. And do the same thing for this one. This is going to simplify to -1,
then we have 2k plus 1 minus 1 over 8k plus 4 plus 3. That's going to be -2k minus 2 minus
1 over 8k plus 7. And again we're going to do the same thing. We're going to multiply by 1
over k and also simplify the values. Negative 2k minus 3, 8k plus 7, and then that equals...If we
multiply by 1 over k, that's going to be -2 minus 3 over k over 8 plus 7 over k. As you can see,
we have different values and those are going to converge pretty obviously to different
numbers. We have a -2 here and we have a 2 here. All right so in this next part,
you're going to prove that the sequences converge and we're going to get real values for the limit
of X of 2k as k goes to infinity and X of 2k plus 1 as k goes to infinity. We have to say
three things before we take the limit. We have to say that we know that the limit of 1 over k goes
to 0. We just know that from prior lessons. We can say that we know that the limit--there's a theorem
that says that this is true--the limit of 1 over k as k goes to infinity equals to 0. And what we're
going to do because we have numerator--we have a numerator and a denominator for both X of 2k and
X of 2k plus 1, we're going to say that because the denominator does not go to 0 in either case,
that the limit of the fraction is the limit of the numerator divided by the limit of the denominator
because they're real values in the denominators. We're going to say that, and we know from algebra,
that 8 is greater than 0 and 7 is greater than--sorry 8 is greater than 0. We're going
to just say that we know that this is 0, so these parts are 0. The limit of the first denominator,
8 plus 3 over k as k goes to infinity, is greater than 0 and the limit of 8 plus 7 over k as k goes
to infinity is greater than 0. Thus we have that. And now we can take the limits. We have
that the limit of X of 2k as k goes to infinity equals the limit of this first value. That's 2
minus 1 over k over 8 plus 3 over k as k goes to infinity. We can break this up now into numerator
and denominator. And because we know that this is 0, we have that this is equal to 2 over 8.
And let's do 2k plus 1 now. We have that k goes to infinity of X of 2k plus 1. That's going to be the
limit of the second one again going to infinity. Again we're breaking that up into numerator and
denominator and that's going to be equal to -2 over 8. We're pretty much done. Now we're just
gonna say at the end because obviously 2 over 8 does not equal -2 over 8, X of 2k and X of 2k
plus 1 do not converge to the same values. Thus X of n is a divergent sequence because
its subsequences do not converge to a single value for all n in the naturals and we're done.