2007-10-26

Today's nonsense
- Homework due Monday. Posted just yesterday on the internet. Due Monday: pg. 628 #3-36
- Email to send out plus link ..
- Spiller was saluditorian, didn't know it until the end of his senior year in high school, lots of racial tension? From Detroit.
- Complaining about the problem on tension on the physics test.
- Spiller hit Ketteman with a ball and yelled "be quiet" (not absuive; not at all).

"Yesterday" we were working on the direct comparison test and now we are going to be continuing. And going on to the:

Limit Comparison Test

Suppose that a_{_n} > 0, b_{_n} > 0, and
limit from n -> infinity of ( a_{_n} / b_{_n} ) = L
where L is finite and positive. Then the two series (Series A_{_n}) and (Series B_{_n}) either both converge or both diverge.

ex 1) Use the limit comparison.test: series from n=1 to infinity of 1 / (2n+3).
- If it's Series of 1/n, then that diverges. So you say it's harmonic, so it diverges, because it's linear, well, let's look at this. The way this works, you're going to take the limit as you approach infinity of the ratio of the two, the one that you know is on the numerator and the one that is similar is on the denominator. What is the limit as n-> infinity of 1/(2n+3) all over (1/n) ... and essentially you look at the largest terms in the numerator and denominator and it's basically n/2n and that's 1/2. The limit comparison test says that if you compare the two functions, and you get a finite value, such as 1/2, then the two series that you are comparing are going to do the same thing. So if one of them you can figure out, and it diverges, then the other one has to diverge, so therefore both diverge in this case because you already know that 1/n diverges. So there you go.

2) Series from n=1 to infinity of (5root(n) + 6) / (2n - 11)
Compare to 1/sqrt(n) ... which diverges. And you can take the limit of the ratio which is something like 5/2. And since 1/sqrt(n) diverges, then our given problem also diverges.

3) Series from n=1 to infinity of 2/(3^n - 5)
Compare to 3^-n where r=1/3 and it's geometric and therefore it converges. The ratio is a finite and positive and nonzero number ... anything else means that it can't converge. You have to compare it with the same power-type function so that they will cancel it out and there will be a ratio of coefficients. And if you do it properly you can figure out that our original given problem (#3) converges.

4) Series from n=1 to infinity of sqrt(n) / (n^2 + 1).
- Compare to 1/( n ^ (3/2) )

5) Series from n=1 to infinity of n2^n / (4n^3 + 1)

27 min