Good morning fellow mathematicians welcome back to my video. I promise this right here is going to be the last
checking for radius of convergence video
for now, so I'm not planning on doing any more of those.
I wanted to talk about this one right here because it's not a special case
but it's probably the first time we're ever going to encounter that you have to check for the boundary points if
this (?) converges
Okay
What do we want to do? Well, first we have to decide for little ra-, um...
convergence test. Convergence test, that's the word. So let's say those are our sequences
a_k. Once again, I'm going to decide for the ratio test because it works out the best
Okay, we are going to have the limit as k approaches infinity of absolute value of a_(k+1)
over a_k and if this thing is strictly less than 1, well then that series converges absolutely.
Ok, here we go. I'm going to refer to this limit as just capital L.
Absolute value of, okay, just plug the k+1 into everything here.
So we're going to get -1 to the k+2 power, k+2
x to the k+1 power over k+1.
A big over over
complex fraction. We are going to get now a_k. So -1 to the (k+1)-th power, x to the k-th power
over k.
A lot of stuff is going to cancel out. The absolute value its multiplicative meaning absolute value of a times b
is the same as the absolute value of a times the absolute value of b (|ab| = |a||b|)
Meaning we can just distribute this absolute value into the -1. The absolute value of -1 is just 1
So let's get rid of this chunk right here.
also x to the k-th power and x to the k-th power is going to cancel out. We are going to get
the limit as k approaches infinity of
Okay. Now we only have x to the first power up here. So let's do mo-, let's use the multiplicative
property of the absolute value to break this up into the absolute value of x times the absolute value of, taking the reciprocal,
k over k+1.
If you don't know where this property right here comes from, just take a look at Papa Flammy's (bless) advent calendar
I made a little proof on that.
k is an element of the natural numbers so this ain't negative never! So that won't be ever negative
so absolute value of something non negative is just
thing itself. And if we take the limit as k approaches infinity, well
we can just factor out the k on both terms right here. So this makes k over k, this is just 1. Plus 1 over k.
This and that is going to cancel out so we are going to have to limit of the absolute value of x
1 over 1 plus 1 over k.
We can just distribute this limit into everything: limit of a constant is just a constant. Also limit of
1 over k as k approaches infinity is just 0 so this overall is nothing but 1.
Meaning we're going to end up with the absolute value of x,
Which we want to be strictly less than 1.
This for now is our
radius of convergence.
But not completely. That's where the special case comes in, you could say.
For all those other sequences I had- for those other series we had
We had the property that our radius of convergence is basically infinite.
So you can't really check for the boundaries of a function where the radius of convergence is infinite.
It really doesn't make much sense. There are no boundary points in this interval.
Also, we had a geometric series, but there was the thing if you sum up infinitely many ones or negative ones
it's just going to diverge into infinity or nowhere at all.
Ok.
But here, how is it about 1 and -1 because absolute value of x strictly less than 1
is equivalent to saying that our x is between -1 and 1.
So we have to check for our boundary points if our
series actually converges or not. Let's plug -1 in and see what we get first. So
Ln of
Okay, I'm not going to put it like this
We are going to write this out.
Okay, we are going to have to sum from k equals to 1 to infinity of
-1 to the (k+1)-th power, -1 to the k-th power
Over k. This makes -1 to the (2k+1)-th power
-1 to some even power just 1 so this makes
Negative the sum running from k equals to 1 to infinity of 1 over k.
And the best thing is, this thing right here is the harmonic series. So this thing is going to converge... NOT.
It's going to diverge, I'm terribly sorry.
It's going to diverge, it's going to go to infinity because you are summing up more and more parts
which don't become any smaller. Not really smaller in the grand scheme of things.
So this thing is going to diverge That's nothing but negative
the harmonic sequence. Um, yeah. I don't know if this has an index or something. Um, never mind. It's going to go to infinity
overall. Ok, so it doesn't diverge- it doesn't converge for -1 but how about +1?
Now we have to sum from k equals to 1 to infinity of -1 to the (k+1)-th power
over k in this case, because 1 to the k-th power is nothing but 1.
Why not write this out a little bit? Okay, at first we get
1 over 1, ok this is just 1.
Plus, ok. This is going to give us negative so
-1 to the third power is negative
1 over 2.
Positive 1 over 3 and it's going to go on. Negative 1/4.
Positive 1/5.
Negative 1/6, and so on.
Why not turn this into something completely
positive, so without those negative science in between here. So
I'm going to leave this first term how it is. It's going to be useful in the middle.
But, on this term we are going to have. Okay, let's take a look at a more general case.
You see we have 1 over n, you could say,
minus
1 over n+1.
Ok, we can advance those fractions: this by n+1 over n+1, so we are going to get n+1 over
n times n+1, and then negative n.
Nothing but; okay this and that is going to cancel out. So this is 1 over n times n+1.
Ok, meaning over all that we are going to get 1 minus 1/2: those two terms.
Then those two are going to be, ok
Plus 1 over what? n times n+1. So 3 times 4 which is nothing but 12.
Plus 1 over 30.
Plus, okay
this is going to be 7 and 8 is going to be 56 and so on and so on.
Now we can actually do a little
approximation now.
If we just switch this 1 and -1/2
and this 1. Well, okay 1 is just 1. That's ok. So this is
less or equal to 1.
1/12 is, well
pretty small, but it's even smaller than
Yeah, 1/4. I hope you agree with me. You can just do a little proof that this holds. Ok
1/30 is even smaller, but it's even even smaller than
1/9 and so on and the next term is going to be strictly smaller than 1/16 and so on.
Meaning, overall this whole sum is
strictly less than negative 1/2 plus 1
plus 1/4 plus
1/9 plus
1/16 plus dot dot dot.
We can rewrite this because this is just, so 1 is nothing but 1 over 1. So 1 is 1 squared.
4 is 2 squared, 9 is 3 squared and so on. We're going to get -1/2
plus the sum running from k equals to 1, in this case,
to infinity of 1 over k squared. And we have evaluated this useful sum before. This is nothing but -1/2
plus π squared over
6.
Not 16, just 6. And you see we have actually found the upper bound
so this obviously is finite meaning this is also finite so
we have actually used a
in German it would be the majorantenkriterium (please check spelling)
Major criteria, I don't know. Leave in the comments what it's called in English. Yeah, meaning our radius of convergence
It's actually like this right here.
on this natural log. And then we are done. Yeah this is kind of a special case, something you might have encountered before in
mathematics when dealing with series. I'm being serious here (punny). It happens sometimes and
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