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Help i'm drowning!

  1. May 17, 2005 #1
    how in the world do i test this series for convergence? I know that i should use the ratio test but i'm not sure how to rewrite the expression:

    [tex]\sum_{n=1}^\infty\frac{(2n)!}{n^n}[/tex]
     
  2. jcsd
  3. May 17, 2005 #2
    I don't know how to solve any of these so i might as well post them.

    [tex]\sum_{n=1}^\infty\frac{\cos{n\pi}}{n}[/tex]


    [tex]\sum_{n=1}^\infty\frac{n^{1000}}{1.001^n}[/tex]


    [tex]\sum_{n=1}^\infty\frac{n!}{1000^nn^{1000}}[/tex]


    [tex]\sum_{n=2}^\infty(-1)^n\frac{n}{2n+1}[/tex]


    all i need to know is if they converge. I don't need to find their sums, if they exist.
    thanks in advance i'm at a loss as to what to do. these are due tomorrow afternoon.
     
  4. May 17, 2005 #3
    these must be some difficult problems! I guess i don't feel so bad now.
     
  5. May 17, 2005 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You are getting no response, not because they are difficult questions (some are, some aren't) but because you have given no indication about what you know, what convergence tests you might try, or what you have actually tried.
     
  6. May 17, 2005 #5

    These are the test that i know.

    1. nth term test for divergence

    2. convergence of p-series

    3. ratio test

    4. integral test

    5. root test

    I have tried what i know. I guess i'm not seeing how to apply these test. I haven't posted here for quick answers because i want to know how to solve these problems myself. I just don't see how to solve them right now.
     
  7. May 17, 2005 #6
    Try the ratio test on the first problem and post how it goes.

    --J
     
  8. May 17, 2005 #7
    i'm not coming up with anything.
     
  9. May 18, 2005 #8
    Clearly, since you're asking the question. I meant post your work and the steps you followed.

    --J
     
  10. May 18, 2005 #9

    for this one i know that i should use the ratio test but i don't know how to simplify the factorial. I know this much

    [tex]\frac{2(n+1)!}{(n+1)^{n+1}}\frac{n^n}{2n!}[/tex]

    is this much correct?
     
  11. May 18, 2005 #10
    this is my fault i've taken too many credits over the summer session. I should know this stuff but with cal III, D.E., world religions, and amh history in 12 weeks it's killing me.
    once i get past tomorrow i can devote about 12-14 hrs to getting caught up.
     
  12. May 18, 2005 #11
    You're close. You might be right, and might've missed a set of parentheses you needed.

    You're looking for

    [tex]\frac{(2(n+1))!}{(n+1)^{n+1}}\frac{n^n}{2n!}[/tex]

    Note that (2n+2)! = (2n+2)(2n+1)(2n)! and that (n+1)^(n+1) = (n+1)(n+1)^n.

    Can you simplify it from there?

    --J
     
  13. May 18, 2005 #12
    Since I need to sleep, I'll give you some hints for the rest.

    For the second one, calculate [itex]\cos{n\pi}[/itex] for a few n. See a pattern? See if you can't represent the cosine as something more familiar, and your series should become familiar.

    Third and fourth can both be determined by the ratio test.

    For the fifth, consider the most basic conditions on [itex]a_n[/itex] for [itex]\sum{a_n}[/itex] to converge.

    --J
     
  14. May 18, 2005 #13
    the third one is inconclusive using the ratio test because the limit of the ratio is equal to 1
     
  15. May 18, 2005 #14


    That is not true. The ratio tends to be infinite (it is compatitable with n).
     
  16. May 18, 2005 #15
    The third one converges and the limit of the ratio is less than 1, it is 1/1.001.

    --J
     
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