2007-10-29
Today's nonsense
- Spiller's levels of pissed-offness, i.e. comparison of Katie v. Avery, which is kind of like a series test because we have to throw one against a wall.
- Video by the science teachers to the song of "Pink Cadillac." The neighbor-teacher took his old 66 Bewic, made fins out of cardboard, painted it pink, and drove around in Blues Brothers attire, with the sunglasses and everything, filmed it, etc. This guy had nothing better to do ... he used to make videos called "Mr. Science's Neighborhood."
Alternating Series
This is where the summation alternates between positive and negative terms in the series. How do we create this alternation? You do (-1)^n. And then you can start with to the power of n+1 and that's an odd number if you're starting at n=1 so that the original negative sign is preserved or not depending on what you're doing. Is that the only way to get alternating? No. You can use sine and cosine.
When is cosine=1? At 0 or at 2pi or 4pi or 6pi. And cosine is -1 at 1 etc.
Series from n=1 to infinity of a_n * cosine (pi * n) = - + - + - + - + ...
Series from n=1 to infinity of a_n * sin( pi/2 + pi * n ) = - + - + - + - + ...
And these are the only the alternating parts of the series that we are going to be dealing with. There are two steps that we are going to work with, and we combine them on the test, to determine diverging or converging.
Alternating Series Test
Let a_n > 0. The alternating series
Series from n=1 to infinity of (-1)^n * a_n
and
Series from n=1 to infinity of (-1)^(n+1) * a_n
converge, if the following two conditions are met.
1. It has to be nonincreasing- so: a_(n+1) < = a_n. Must stay the same or decrease.
2. n^th term test. If the limit as n-> of a_n = 0. This has nothing to do with the alternating sign, so the absolute value of the sequence has to be decreasing or staying the same.
Does it converge or diverge?
ex 1) Series from n=1 to infinity of (-1)^(n+1) * (1/n)
Can we just write converges or diverges? No. We have to acknowledge this stuff. So we have the alternating series test
Is a_n > = a_(n+1) ? Check. So you have to check if 1/n > = of 1/(n+1) ... and then you can see that n+1 > = n. And 1 > = 0, so yes, the first one works.
-- And does the limit as n->infinity of 1/n = 0? Check.
It converges by the alternating series test.
2) Series from n=1 to infinity of n / ( (-2)^(n-1) ) .......
So now plug this stuff in to the a_(n+1) < = a_n requirements and see if it qualifies. You can rewrite this series as (-1)^(n-1) * n/(2^(n-1)). So how about the alternating series tests? Can we check? Yes, checking is good. Can we try to verify that the first condition is met? Is 2n > = n + 1? Yes. As long as we are starting with n=1 or starting with n > 1. We have proved condition 1. So now we have to work with condition two ...
Then do the n^th term test. So what about the limit as n -> infinity of n / 2^(n-1) and the demonitor grows much faster than the numerator. So therefore it goes to 0. And so therefore this series converges by the alternating series test.
If it's an alternating series, now, alternating series means that every other term, so you can't go positive positive negative negative positive negative; every term is the opposite of the last one and the next one. If these two conditions of an alternating series are met, then it converges.