2007-10-16 - Introduction to series

Today's nonsense
- Today's lesson is dedicated to Bobby who rocks so there must also be paper and scizzors in here somewhere. First he had email troubles yesterday, and today he wasn't going to come because he's sick, but now he's here.
- Avery claims it is AntiAvery Day again, but that was clearly yesterday. What about the song that says that "short people have no reason to live," is that how it goes?
- Bingo next week.
- Math club on Monday morning? Student council is at that time. NHS is every other Tuesday.
- Homework due.
- The story of cornell and it starting to rain corn? (What's with this?)
- "The only constant in life is change." Or Luisa's version: death.
- Do lakes have to have flowing water into it? Or are these only man-made lakes? (Katie's tangent ... blame her?)
- What was Spiller doing growing up in a Polish neighborhood?

A sequence is a number of terms in a series. So we're going to talk about series today, tagging.

Infinite series
sum from n=1 to infinity of a_{_n} = a_1 + a_2 + a_3 + a_4 + ... + a_n + ...
These series go to infinity. So the question will be do the series, which are sigma which is summation, do they converge to a specific value or not? When you add up all the terms of the sequence, do they converge to a specific sum or not? Which are convergent and which aren't?

The terms have to be getting smaller for it to be convergent. Do they have to be monotonic? If it's monotonic and bounded, then it converges, but that's the sequence, so what about the series? Let's look at some partial sums.

Partial Sums

A partial sum means a sum of a finite number of terms. The first partial sum would be nothing but the first term of the sequence. The second partial sum would mean add the first two terms. So they might say, "estimate the sum of the series by using a seventh partial sum" meaning add the first seven terms, and so then you get an estimation of what the total sum is.

So in partial sums, what you do is you take the number of terms up to that nth partial sum, and find what that is, and after a few of these partial sums, you can start to look for a pattern, and then take the limit of that pattern to see if it converges. You have to be able to prove that a series converges to a particular number.

ex 2) the sum from n=1 to infinity of (1/n) - 1/(n+1)
- A telescoping series

ex 3) sum from n=1 to infinity of 2/(4n^2 - 1)
... Even though we are already using partial sums, we have to use partial fractions and this is because it's a telescoping series. We're going to want to call this "A over" and "B over" and what do those factor into? How do you factor 4n^2 - 1 ? That's 2n-1 and 2n+1. Our original problem can be rewritten as the same sum of 1/(2n-1) - 1/(2n+1) and this is in the form of the telescoping series. So now you have to find the sum, just say what the answer will be, because everything else drops out. List the first few terms and see what it is going to add up to be. So this is where you have the form where you have something canceling out, and quite often you will have to do partial fractions.

Now something that you dealt with in precalculus

Geometric series

Arithmetic: adding the numbers
Geometric series: where you are multiplying the terms.

Sum from n=0 to infinity of ar^{^n} = a +
When n=0, what is the first term? In this case, each time in the geometric series, whatever is raised to the n, is the "r" or your common ratio. What is ar^2 divided by ar? It's r, and that's your common ratio. The same works for dividing the fourth by the third terms. This is a geometric series with ratio r.

Convergence of a geometric series

If you keep multiplying something by a number greater than one, is that thing ever going to converge? No. What if the r-value is 1? No, it'll be the same still. There has to be some kind of restriction on r in order to have a converging geometric series.

The geometric series diverges if |r| >= 1 so that means if |r| is below 1 then it will converge.
To converge, 0 <= |r| <= 1. The r-value can be negative (meaning the terms are alternating). In the case of 0 <= |r| <= 1, the geometric series converges to the sum (see below)

sum from n=0 to infinity of ar^{^n} = a / (1-r) ---> and 0 <= |r| <= 1

We can prove that easily. Sometimes this is completely misunderstood ... some examples will now follow, to try to throw out that misunderstanding:

ex) Determine convergence/divergence. Find sum if convergent.

a) sum from n=0 to infinity of 3 (1/4)^n
Use the form that was given for the geometric series.

b) sum from n=0 to infinity of 4(5/4)^n

c) sum from n=1 to infinity of (2/3)(1/3)^n =

Paul's online math notes - convergence of series
Convergence tests for infinite series
BOX!
Videos - (Selwyn Hollis, 2005) (University of Houston)