Calculus Two: Sequences and Series is an introduction to sequences, infinite series, convergence tests, and Taylor series. The course emphasizes not just getting answers, but asking the question "why is this true?"

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From the course by The Ohio State University

Calculus Two: Sequences and Series

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Calculus Two: Sequences and Series is an introduction to sequences, infinite series, convergence tests, and Taylor series. The course emphasizes not just getting answers, but asking the question "why is this true?"

From the lesson

Convergence Tests

In this third module, we study various convergence tests to determine whether or not a series converges: in particular, we will consider the ratio test, the root test, and the integral test.

- Jim Fowler, PhDProfessor

Mathematics

P series.

[SOUND].

[MUSIC].

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

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

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

The claim is the following. This series converges

if p is digger than 1.

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

We can check this using Cauchy Condensation.

Well lets just remember what Cauchy Condensation says.

It says to analyze the convergence of a given series,

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

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

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

How did I get this?

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

So now, all I've got to do is

just determine whether this condensed series converges or diverges.

How am I going to do that?

Well this series can be rewritten.

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

1 power.

Look, I got 2 to the n in the numerator

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

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

Now this I could also rewrite.

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

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

Why is this an improvement?

What kind of series is this?

Right?

This series is just a geometric series with common

ratio 1 over 2 to the p minus 1.

And by condensation, if this series converges or

diverges, so too does the original p series.

So I've reduced the entire question down to just

knowing whether or not this geometric series converges or diverges.

And I know how to analyse that.

If p

is bigger than 1, then this common ratio is

less than 1 and in that case the series converges.

And if p is less than or equal to 1,

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

And in that case the series diverges.

If you have already seen integrals, you could

also do this by using the integral test.

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

to the question about this integral.

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

this series converges and that happens if and only if the

integral from 1 to infinity of 1 over x to

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

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

over x to the p, is positive and

decreasing on the interval that I care about.

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

just analyse this integral and if this

integral has a finite value, this series converges.

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

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

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

definition, this is the limit as n approaches infinite

of the integral from 1 to big N of

1 over x to the p, dx.

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

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

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

Now I want to evaluate this.

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

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

things happened right of p is, is 1.

But we've already handle the case were p

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

Okay, so lets keep going assuming that p

is not 1, then I can evaluate this definite

integral by finding an anti-derivative of x to the

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

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

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

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

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

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

So this is the limit begin, goes to

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

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

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

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

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

This limit has a finite

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

Well how do I check that?

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

N to the minus p plus 1. What happens?

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

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

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

So then the series diverges.

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

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

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

this series converges.

So then converges and maybe it's a little bit

complicated to think about what these two conditions are.

Instead of saying minus p plus 1 is positive,

I could instead just say p is less than 1.

And instead of saying minus plus 1 is negative,

I could just say that p is bigger than 1.

So now we've shown the whole

story right.

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

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

Even when the P series converges, the actual

values that we are getting can be quite mysterious.

For example the sum of 1 over n squared

and goes 1 to infinity is pie squared over 6.

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

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

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

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

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

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

Well, hundreds of years ago, Euler calculated

an approximation.

He calculated some of the decimal digits of this number.

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

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

He gets stories pretty cool, right?

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

And over

200 years later, Apery comes along and says

that number that you were approximating is irrational.

I mean, to be doing mathematics means that

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

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

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