namely, 1/(1-𝑥)².
I know.
So, here we go, let's just put it down right here.
First of all, let's write down what we know-- of course, our Best Friend, 1/(1-𝑥).
This right here is the sum (Σ) as 𝑛 goes from 0 to infinity (∞), 𝑥ⁿ,
and don't forget that we should also give the radius and also the interval of convergence,
so I will tell you based on this, we know |𝑥|<1, and, with that being said, we know 𝑅 is 1,
and from here, we can also say the 𝐼-- for the Best Friend, we go from -1 to 1,
and we do not include both endpoints.
So, this right here is what we know.
Now, check this out.
How can we go from 1/(1-𝑥) to 1/(1-x)²?
Well, you might be thinking:
"Can't we just square both sides?"
Technically, yes, but it's going to be harder to do,
because when you square a power series right here, remember, you have to put the parenthesis here,
and then square.
And remember, this right here is an infinite polynomial; when you square that, it's not easy to do.
It's doable, but it's not easy, all right?
But anyway,
Hmm... let's think about how we can... do another operation
so that we can go from our Best Friend to this new friend.
Ah! Don't forget about calculus; of course, we're in a Calculus class, right?
If we integrate our Best Friend, we'll end up with some kind of natural log (short for logarithm), right?
How about-- if we differentiate our Best Friend, what will we get? Right?
Well, if we differentiate this guy, based on the Kuoshen Lu (Quotient Rule), if you would like,
you will see that you'll have to square the denominator, and that's the good thing.
So, this is what we'll do:
We'll just look at this and differentiate both sides and see what happens.
So, here we go.
Based on this, we are going to differentiate both sides.
Now, perhaps I'll write this down here for you guys, real quick, somewhere on the side.
Let me just put down some note right here, for you guys.
Perhaps the easiest way to differentiate this is this:
I will put this down.
When you differentiate that, I can look at this as (1-𝑥)⁻¹, like this.
And, when we do that, we can bring the power to the front and then minus 1,
so we get negative, and then this right here stays the same, so we have 1-𝑥, raised to the -2 power.
But don't forget the Chen Lu (Chain Rule); we have to multiply by the derivative of the inside.
The derivative of 1-𝑥 is -1, so be sure that you multiply by -1.
This times that is positive; this right here can go down to the denominator,
so, all in all, we get 1/(1-𝑥)².
Very, very nice.
Therefore, on the left-hand side, we get nicely equal to 1/(1-𝑥)². Very nice.
And, as of the right-hand side,
of course, we can just do our usual derivative with the power right here,
because this is technically a power function.
We can just bring the 𝑛 to the front, and then minus 1. That's pretty much it.
So, we can say this right here is equal to Σ as 𝑛 goes from something to ∞---
you'll see why I said "something", all right?
Well, you can put the 𝑛 right here first,
𝑛 and then 𝑥ⁿ⁻¹, like this, right?
So now, here's the deal:
earlier, it was 0; if I maintain the 0, it's okay, it's not wrong, but check this out.
If I put 𝑛=0, well, the first term is, you plug in this 𝑛 and that 𝑛,
you will end up with 0𝑥⁻¹, like that.
0 times this is just 0, so it's nothing.
Usually, when these kind of things happen, we would just like to say 𝑛=1.
We lose the first term because the first term was a constant.
So, when you have 𝑛=1, you know, you can start with the actual term, the non-zero term.
But anyway, this right here is the form of the power series for this guy, 1/(1-𝑥)².
This is, like, our Best Friend, and you power that up, right?
Hmm... let's see, what else can we do right here?
I think we also need to get the 𝑅 and also the 𝐼, so let's check that out.
Well, when we differentiate, the good thing is that the radius of convergence will stay the same,
but we should double-check the interval of convergence.
So 𝐼 is going to be from -1 to 1.
But let's just do this real quick.
Well, check the radius-- the convergence at the endpoints,
so I will check when 𝑥=-1, we put this right here,
and we will get Σ as 𝑛 goes from 1 to infinity,
and, again this is just checking for the convergence at the endpoints.
Putting -1 into this 𝑥, we get 𝑛(-1)ⁿ⁻¹.
This right here is alternating, but this is 𝑛 times that.
Unfortunately, this doesn't even go to 0 in the first place.
If you look at this-- if you take the limit as 𝑛 goes to ∞ of 𝑛(-1)ⁿ⁻¹,
this right here is technically is not-- well, D.N.E. A doesn't-exist situation,
And "doesn't exist", of course, does not mean 0, therefore, this right here diverges,
by T.F.D. (Test For Divergence)
So, this right here, we still do not include the endpoint, right? The endpoint is right there.
And now, we'll have to check 𝑥=1.
Put 1 into the 𝑥 right here, so we get Σ as 𝑛 goes from 1 to ∞, 𝑛(1)ⁿ⁻¹,
this is just always 1, in other words, this is Σ as 𝑛 goes from 1 to ∞ namely, all of the natural numbers,
namely, 1, 2, 3, 4, 5, 6, and so on, and you just add them up,
the answer to that is not -1/12; if you know the reference, leave a comment down below,
But anyway, this right here, I'm going to tell you if you take the limit of this as 𝑛 goes to ∞,
you technically would end up with ∞, and, of course, ∞≠0.
This right here, technically, it's not ∞ because here you have -1 and then positive 1,
so it kind of bounces up and down,
so the answer to this limit is "Doesn't Exist",
but the answer to this limit is ∞.
But anyway,
this is also not 0, so we get a divergent when 𝑥=1,
and this diverges by Test For Divergence.
Come here and give that a parenthesis.
and with all of that, we are done.
Uh, I'll just box everything right here for the answer, like this.