Diary outline for Math 152-pt2,

# Diary outline for Math 152-pt2, - Diary outline for Math...

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Math 152 diary, spring 2009: second section Later material Previous material In reverse order: the most recent material is first. Monday, April 13 (Lecture #21) About the exam, etc. We'll have an exam next Monday, April 20. Here is some review material. Here is an outline of (possibly!) useful approaches to sequence/series problems. On Thursday, April 16, students will work on the course coordinator's review problems. I will not be available this Thursday and Friday (out of town for a professional trip) but I will have a review session for the second exam on Sunday, April 19, from 5 to 7 PM. The location for this is SEC 117. Why the Root Test? The Root Test is another method for exploiting similarity with geometric series to diagnose absolute convergence (or divergence) of a series. We consider a series ∑a n . Suppose we think that a n "resembles" cr n . Well, if we take the n th root of cr n , we get (cr n ) 1/n =c 1/n (r n ) 1/n =c 1/n r n·(1/n) =c 1/n r. Now if n→∞, we have already seen that a sequence like {c 1/n } has limit 1. So (cr n ) 1/n →r as n→∞. So we can hope that the asymptotic behavior of (a n ) 1/n as n→∞ can help analyze the convergence of ∑a n . Statement of the Root Test The Root Test Consider the series ∑ n=0 a n , and the limit lim n→∞ |a n | 1/n . Suppose that this limit exists, and call its value, L. (L is what's used in our textbook.) If L<1, the series converges absolutely and therefore must converge. If L>1, the series diverges. If L=1, the Root Test supplies no information Ludicrous example Let's consider the series ∑ n=1 ((5n+7)/(6n+8)) n which I invented specifically to use the Root Test. I don't know any application where I've ever seen anything like this series which seems fairly silly to me. Well, anyway, the terms are all positive, so I can "forget" the absolute value signs. We take the n th root and remember that repeated exponentiations multiply: [ ((5n+7)/(6n+8)) n ] 1/n =((5n+7)/(6n+8)) n·(1/n) =((5n+7)/(6n+8)) 1 =((5n+7)/(6n+8)). Now we need to discover the limit (if it exists!) of ((5n+7)/(6n+8)). But (L'Hôpital) this limit is 5/6. Since L=5/6<1, the series converges absolutely and must converge. I don't know what the sum is. Oh well. Less silly (maybe) example This may look almost as ludicrous, but it turns out to be more significant. Again, though, this example is chosen to work well with the Root Test. For which x's does the series ∑ n=1 n n x n converge? The powers of n signal to me that probably I should try the Root Test. Here a n is n n x n . We can't just discard the absolute value here, but we can push it through to the x because everything is multiplied . So: |n n x n | 1/n =(n n |x| n ) 1/n =n n·(1/n) |x| n·(1/n) =n|x|. As I mentioned in class, as n→∞ jumping to the "conclusion" may be unwise. There are

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Diary outline for Math 152-pt2, - Diary outline for Math...

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