The most interesting series in the world.
[SOUND]
[MUSIC]
Power series don't always converge.
But when they do, they usually converge absolutely.
Here's the theorum.
Suppose that this power series converges when I plug in x nought x sub 0.
Well, then that same series converges absolutely for any value of
x between negative x nought and x nought. Well, let's prove the theorem.
Well, the assumption is
that the power series converges when I plug in x nought.
And a consequence of conversions is that the limit of the nth term, which
in this case is a sub n times x nought to the nth power.
Well, that limit must be 0 because this 0 is conversion when I plug x nought but a
conversion sequence is a bounded sequence. What that means
is that there is an m, so that
for all n, this is no bigger than
m in absolute value. So I'll write that down.
a sub n, x naught to the nth power is less than or equal to m.
Now, pick an x. Well, I'm going to pick that x to
be between minus x nought and x nought. So I'll just write that here.
I'll pick some value of x that's in the interval
between minus the absolute value of x nought and x naught.
So this just means x is an element of this interval between minus
the absolute value of x naught and the absolute value of x naught.
Writing it in this funny way just because x nought might be negative.
My goal is to show that with that value of x.
The power series converges absolutely.
Now, watch this.
The absolute value of an sub n times x to the nth power, well it just
equals to the absolute value of a sub n times x nought to the nth power.
Times the absolute value of x to the n, over x nought to the n.
I mean this equality is just cause x nought to the n dividing by x
naught to the n, got an x to the n here, and these are just equal.
But, what else do I know?
I know that this part here, sub n times x nought to the nth power is
bounded by big m. So this is less than or equal to big m
and I could rewrite this as the absolute value of x over x nought to the
nth power. But that helps me make a comparison.
Well, how so?
Well let's set r equal to the absolute value of x over x naught.
And then let's think about this series, the sum n goes from
0 to infinity of m times m times r to the n.
So, it's exactly the turn here but I'm setting them all up.
That series converges. Well,
since x was chosen to be between negative axis
value of x not and absolute value of x not.
This quality all the ratio between x and x nought, and absolute value that is less
than 1, and consequently this just geometric series, well it converges.
So what about the original series? So by the direct comparison test,
what do I know?
I know that the sum n goes from 0 to infinity of the.
Absolute value of a sub n times x to the n converges.
Original power series converges absolutely at that value of x.
And remember back to what the original statement of the theorem was.
I'm just assuming convergence at a point and then I'm deducing
Something much stronger, absolute convergence on a whole interval.
It's an important corollary of this theorem.
So consider this power series, then there is an
r, so that series converges absolutely for any value of
x between negative r and r and it diverges
whenever the absolute value of x is bigger than r.
Meaning whenever x is bigger than r or x is smaller than negative r.
Now I'm not saying anything about what happens when x is equal to r or when x is
equal to negative r. But atleast on this interval, I'm getting
absolute convergents that read our has a name.
This r is called the radius of convergence.
[SOUND]