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The most interesting series in the world.

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[MUSIC]

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Power series don't always converge.

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But when they do, they usually converge
absolutely.

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Here's the theorum.

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Suppose that this power series converges
when I plug in x nought x sub 0.

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Well, then that same series converges
absolutely for any value of

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x between negative x nought and x nought.
Well, let's prove the theorem.

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Well, the assumption is

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that the power series converges when I
plug in x nought.

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And a consequence of conversions is that
the limit of the nth term, which

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in this case is a sub n times x nought to
the nth power.

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Well, that limit must be 0 because this 0
is conversion when I plug x nought but a

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conversion sequence is a bounded sequence.
What that means

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is that there is an m, so that

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for all n, this is no bigger than

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m in absolute value.
So I'll write that down.

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a sub n, x naught to the nth power is less
than or equal to m.

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Now, pick an x.
Well, I'm going to pick that x to

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be between minus x nought and x nought.
So I'll just write that here.

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I'll pick some value of x that's in the
interval

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between minus the absolute value of x
nought and x naught.

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So this just means x is an element of this
interval between minus

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the absolute value of x naught and the
absolute value of x naught.

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Writing it in this funny way just because
x nought might be negative.

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My goal is to show that with that value of
x.

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The power series converges absolutely.

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Now, watch this.

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The absolute value of an sub n times x to
the nth power, well it just

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equals to the absolute value of a sub n
times x nought to the nth power.

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Times the absolute value of x to the n,
over x nought to the n.

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I mean this equality is just cause x
nought to the n dividing by x

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naught to the n, got an x to the n here,
and these are just equal.

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But, what else do I know?

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I know that this part here, sub n times x
nought to the nth power is

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bounded by big m.
So this is less than or equal to big m

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and I could rewrite this as the absolute
value of x over x nought to the

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nth power.
But that helps me make a comparison.

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Well, how so?

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Well let's set r equal to the absolute
value of x over x naught.

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And then let's think about this series,
the sum n goes from

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0 to infinity of m times m times r to the
n.

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So, it's exactly the turn here but I'm
setting them all up.

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That series converges.
Well,

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since x was chosen to be between negative
axis

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value of x not and absolute value of x
not.

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This quality all the ratio between x and x
nought, and absolute value that is less

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than 1, and consequently this just
geometric series, well it converges.

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So what about the original series?
So by the direct comparison test,

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what do I know?

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I know that the sum n goes from 0 to
infinity of the.

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Absolute value of a sub n times x to the n
converges.

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Original power series converges absolutely
at that value of x.

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And remember back to what the original
statement of the theorem was.

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I'm just assuming convergence at a point
and then I'm deducing

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Something much stronger, absolute
convergence on a whole interval.

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It's an important corollary of this
theorem.

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So consider this power series, then there
is an

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r, so that series converges absolutely for
any value of

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x between negative r and r and it diverges

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whenever the absolute value of x is bigger
than r.

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Meaning whenever x is bigger than r or x
is smaller than negative r.

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Now I'm not saying anything about what
happens when x is equal to r or when x is

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equal to negative r.
But atleast on this interval, I'm getting

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absolute convergents that read our has a
name.

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This r is called the radius of
convergence.

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