2007-11-02
Today's nonsense
- No bells because the administrators screwed up. Surprise?
- Homework due Tuesday.
1. Converges, Alt Series Test <-- you must tell whether or not it's conditional or absolute
2. Converges, nth Term Test <-- you can't have a convergence by an nth term test. If it's zero, you have to try another test.
3.Converges, Direct Comparions Test <-- What did you compare it to?
#7 Series from n=1 to infinity of sin(1/n) ... you were supposed to use the limit comparison to 1/n, and that gives you zero over zero, use L'Hopitals rule, and it is equal to 1, so it diverges. The sequence converges to 1, but the series diverges.
#10 Series from n=0 to infinity of (3^n + 2) / (2^n - 3)^2 .... which converges to 3^n / 4^n and that's geometric with r < 1 therefore it converges.
pg 645 #13-30, 37-70
Ratio Test
1. Series a_n converges absolutely if lim as n->infinity of |a_(n+1) / a_n | < 1. So this is just a geometric ratio where r is less than 1 and it converges.
2. Series a_n diverges if the lim as n->infinity of |a_(n+1) / a_n | > 1.
3. The Ratio Test is inconclusive if the lim as n->infinity of |a_(n+1) / a_n | = 1.
- So go try something else. Sorry, the ratio test couldn't help you today.
- See photograph of the Box.
ex 1) Determine convergence or divergence of Series from n=>0 to infinity of 2^n / n!
So take the limit as n->infinity of | ( 2^(n+1) / (n+1)! ) * n! / 2^n | = lim as n->infinity of | 2 / (n+1) | = 0 and that's less than 1, yes, so, this thing (series) converges.
- We multiplied by the reciprocal instead of just dividing like the Ratio Test box suggests.
2) Series from n=0 to infinity of n^2 * 2^(n+1) / 3^n
- The limit is (2/3). So this thing converges by the ratio test.
3) Series from n=1 to infinity of n^n / n!
- Use the ratio test when you have a factorial or an "n" exponent, so this is a good candidate to use the ratio test on.
- The lim as n->infinity of | (n+1)^(n+1) / (n+1)! * n!/n^n | .... and this is a really, really really good one to simplify.
lim as n->infinity of | (n+1)^(n+1) / (n+1) * 1/n^n |
lim as n->infinity of | (n+1)^n / n^n | = lim as n->infinity of | ((n+1)/n )^n |
lim as n->infinity of | (1+ 1/n)^n | = e>1 So it diverges because e is greater than 1.
Later we will do more with the ratio test. After the test.
4) Series from n=1 to infinity of (-1)^n * sqrt(n) / (n+1)
- Recognize that this is an alternating series too. But we'll be using the ratio test.
---> lim as n->infinity of | sqrt(n+1) / (n+2) * (n+1) / sqrt(n) | =
And that's basically n^(3/2) to the n^(3/2) ... so it's practically 1 and it's inconclusive, so moving on ...
- By AST it converges conditionally, because 1/n^(1/2) by the p-series, 1/2 is less than 1, so ...
-- Also look up the root test.
If you take away the power, you take the nth root, then the thing you have left has to be less than 1, and that's sort of what we've been talking about all along.