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Testing for divergence

  1. Mar 16, 2007 #1
    1. The problem statement, all variables and given/known data

    Does the sum of the series from n=1 to infinity of 1+sin(n)/10^n converge or diverge.

    2. Relevant equations

    3. The attempt at a solution

    I can use the comparison test or the limit comparison test.
    I'm not sure where to go from here.
  2. jcsd
  3. Mar 16, 2007 #2
    What can you tell me of the limit of the series as n reach infinity?
  4. Mar 16, 2007 #3
    Well, the top part diverges, the bottom causes it to go to 0. So I don't know what happens faster.

    Either it converges to 0, or it diverges.

    The solution must involve the comparison test or the limit comparison test. But I'm not sure what to compare it to.
  5. Mar 16, 2007 #4
    is the limit of the series as n goes to infinity is not 0 then the sum of the series diverge...
  6. Mar 16, 2007 #5
    wait is it (1+sin(n))/10^n or 1+ (sin(n)/10^n)?
  7. Mar 16, 2007 #6
    Try comparing sin n to n
  8. Mar 16, 2007 #7
    if it's (1+sin(n))/10^n then can you tell me 1+sin(n) is smaller then what for all n?
  9. Mar 16, 2007 #8
    It's (1+sin(n)). Hrm, smaller than 2. So I can compare it to 1/5^n. Now, I need to figure out how to prove that series converges. Is it a geometric series?

    Actually, I know it converges, based on the root test. But I don't think we can use the root test now.
    Last edited: Mar 16, 2007
  10. Mar 16, 2007 #9
    right but 1/5^n is wrong, keep it 2/10^n, now can you tell me if you know the root or the ratio test of a series?
  11. Mar 17, 2007 #10
    alright. So root test gives me limit of 2^1/n / 10. I don't know what 2 ^1/n goes to. Is that even possible?
  12. Mar 17, 2007 #11
    The root test is for when n goes to infinity..
    1/n~0--->2^1/n=1,so (2^1/n)/10<1
    you have just now proved that the series 2/10^n converge, how can you relate this to the series you started with?
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