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Direct comparison test, just need some explanation

  1. Jul 29, 2011 #1
    2 problems, I need to use the direct comparison test with either a p series or a geometric series
    1)series of j^2/(j^3 +4j +3)
    I thought of comparing it to j^2/J^3 which comes out to 1/j, but that dosent work, my teachers answer is you compare it to 1/5j
    2) series of sqrt(q)/(q+2) I would figure it would be 1/q^(1/2), but nope that dosent work, my teachers answer is comapre it to 1/3q^(1/2)

    Can you explain to me how he got those? i feel like all he did was multiply the denominator by some arbitrary number to get an answer that suited his needs? is there more to this?
     
  2. jcsd
  3. Jul 29, 2011 #2
    1.

    In order to satisfy the conditions of the comparison test, if you want to show your original series diverges, you need something smaller or equal to. You can *guarantee* that the series you are comparing it to will be smaller by taking something with a larger coefficient than the 4j, such as.... 5j. Basically if you chose 1/j, the series would *not* be smaller for all n>0. So by choosing 5j, you satisfy the conditions of the comparison test.

    2. Again, this is an issue of thinking "If I compare it to series b, will series b always be smaller than series a?" The series you are using to compare it to will *not* be smaller unless you take your series to be at least 1/3j^(1/2)


    What I like to do is when you are faced with something simple along these lines, test a few values of n such as n = 1, n = 2, and see if the series you are trying to use will meet the requirements for the comparison test. For instance in 2. if you take n=1, your comparison b would be 1, but the original series would be 1/3, so your series would *not* be smaller. Since you need it to be equal to or smaller, make your comparison series something that will be at least 1/3 at n=1 and still meet the requirements of the test.

    Edit: I suppose it would be somewhat arbitrary to meet your needs of the test, but that is the beauty of the comparison test. You could have chosen for the first one 1/(100000 * j) and still have been correct in it's usage.
     
    Last edited: Jul 29, 2011
  4. Jul 29, 2011 #3

    Dick

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    Yes, there's a little more to it. It's true j^2/(j^3+4j+3) is basically like 1/j. But you know the sum 1/j DIVERGES. To prove j^2/(j^3+4j+3) diverges you need a comparison series that diverges that is LESS than j^2/(j^3+4j+3). The problem with using 1/j is that 1/j is GREATER than j^2/(j^3+4j+3). Showing a series is less than a divergent series is no help. That's what the extra factors are about.
     
  5. Jul 30, 2011 #4
    If you start with 4j3 > 4j + 3, this implies 5j3 > j3 + 4j + 3 (for j > 2).
    Can you see how using this comparison really just comes down to what your teacher said with comparing to 1/5j?
     
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