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P series.

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[SOUND].

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[MUSIC].

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When I say P series, this is what I mean.
Well I mean the series, sum n goes

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from 1 to infinity of 1 over n to the p.
For some fixed real number p.

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The question, then, is for which p does
the series converge.

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The claim is the following.
This series converges

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if p is digger than 1.

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And it diverges, if p is less than or
equal to 1.

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We can check this using Cauchy
Condensation.

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Well lets just remember what Cauchy
Condensation says.

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It says to analyze the convergence of a
given series,

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it's enough to look at the convergence of
the condensed series.

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In this case, I'm trying to analyse the
conversion to

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this p series.
And here's the condensed series.

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How did I get this?

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Well, this is 2 to the n times the 2 to to
the nth term of the original series, okay.

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So now, all I've got to do is

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just determine whether this condensed
series converges or diverges.

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How am I going to do that?

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Well this series can be rewritten.

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This series is the same as just 1 over 2
to the n to the p minus

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1 power.

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Look, I got 2 to the n in the numerator

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and a power of 2 to the n in the
denominator.

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So I can rewrite that power in the
denominator, as just the p minus 1 power.

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Now this I could also rewrite.

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I could rewrite this as the sum n goes 0
to infinity

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of 1 over 2 to the p minus 1 to the nth
power.

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Why is this an improvement?

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What kind of series is this?

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Right?

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This series is just a geometric series
with common

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ratio 1 over 2 to the p minus 1.

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And by condensation, if this series
converges or

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diverges, so too does the original p
series.

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So I've reduced the entire question down
to just

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knowing whether or not this geometric
series converges or diverges.

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And I know how to analyse that.

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If p

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is bigger than 1, then this common ratio
is

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less than 1 and in that case the series
converges.

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And if p is less than or equal to 1,

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then this common ratio is bigger than or
equal to 1.

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And in that case the series diverges.

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If you have already seen integrals, you
could

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also do this by using the integral test.

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So the integral test reduces the question
about this series is convergence,

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to the question about this integral.

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In this case, I'm trying to determine
whether or not

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this series converges and that happens if
and only if the

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integral from 1 to infinity of 1 over x to

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the p, dx, has a finite value, if this
integral converges.

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And this is true in this case, because
this function f of x equals 1

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over x to the p, is positive and

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decreasing on the interval that I care
about.

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Okay, so all I've going to do now is

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just analyse this integral and if this

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integral has a finite value, this series
converges.

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If this integral ends of being infinity,
this series diverges.

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So let's see if I can compute this
integral.

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So I'm trying to integrate from 1 to
infinity, 1 over x to the p, dx, and by

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definition, this is the limit as n
approaches infinite

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of the integral from 1 to big N of

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1 over x to the p, dx.

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And now, instead of writing 1 over x to
the p, I can write x to the minus p.

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So this is the limit, n goes to infinity,
the integral

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from 1 to big N of x to the minus p, dx.

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Now I want to evaluate this.

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So, I really want to assume that p isn't
1, because

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I am going to write down an
anti-derivative of this and different

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things happened right of p is, is 1.

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But we've already handle the case were p

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is 1 anyhow, that's just a harmonic
series.

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Okay, so lets keep going assuming that p

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is not 1, then I can evaluate this
definite

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integral by finding an anti-derivative of
x to the

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minus p, right, via the fundamental
theorem of calculus.

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So this is the limit and goes to infinity,
what's an anti-derivative of x

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to the minus p, x to the minus p plus 1,
over minus p plus

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1, all right, then write it.
Add 1 to that exponent and I divide by

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that same quantity.
And I'm evaluating this, from 1 to n.

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So now I've just to plug in n and plug in
1 and take the difference.

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So this is the limit begin, goes to

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infinity of n to the minus p plus 1 over
minus p plus 1.

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Minus what I get when I plug in a 1, which
is 1 over minus p plus 1.

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I could combine these these two fractions,
this is the limit and goes to infinity

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of n to the minus p plus 1 minus 1 over
minus p plus 1.

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Okay, now the whole story is being told by
this limit.

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This limit has a finite

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value, then so does this integral and that
means that this series converges.

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Well how do I check that?

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Well the story is really being told by
this power here, right?

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N to the minus p plus 1.
What happens?

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So, if minus p plus 1 is positive, then n
to

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a positive power, this is going to be very
large and well

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then this limit then will be infinity.
That means that the series will diverge.

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So then the series diverges.

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If, on the other hand, minus p plus 1 is
negative and I'm taking a big number,

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but I'm raising it to a negative power.
It's going to be ending up close to 0.

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And in that case, this integral will have
a finite value and that means that

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this series converges.

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So then converges and maybe it's a little
bit

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complicated to think about what these two
conditions are.

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Instead of saying minus p plus 1 is
positive,

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I could instead just say p is less than 1.

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And instead of saying minus plus 1 is
negative,

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I could just say that p is bigger than 1.

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So now we've shown the whole

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story right.

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If p is less than 1 or equal to 1, the
series

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diverges and if p is greater than 1, then
the P series converges.

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Even when the P series converges, the
actual

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values that we are getting can be quite
mysterious.

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For example the sum of 1 over n squared

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and goes 1 to infinity is pie squared over
6.

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The sum of the reciprocals to the 4th
powers, the sum of 1 over

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n to the 4th, n goes 1 to infinity is pie
to the 4th over 90.

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The sum of the reciprocals of the 6th
powers, the sum the

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goes from 1 to infinity of 1 over the n of
the 6th.

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It's pie to the 6th over 945 so this is 2,
4, 6 we'll go to odd numbers,

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all right, what's the sum of 1 over n
cubed, n goes from 1 to infinity.

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Well, hundreds of years ago, Euler
calculated

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an approximation.

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He calculated some of the decimal digits
of this number.

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And then not so long ago, Apery, in 1978,
showed that this was an irrational number.

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But as far as we know, it's not a rational
multiple of pie cubed.

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He gets stories pretty cool, right?

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I mean, in the 1730s, Euler approximates
some number.

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And over

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200 years later, Apery comes along and
says

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that number that you were approximating is
irrational.

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I mean, to be doing mathematics means that

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you're part of a community that will
outlive you.

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You're joining into a conversation with
people from hundreds of years ago.

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[SOUND]