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Determine whether the sequence converges or diverges

  1. Sep 14, 2011 #1
    Hey all,

    University for me started last week, however I was unable to attend until now. I just e-mailed my professor and there is a quiz just next week and I have no notes and he will not give them to me. I'm wondering if anyone knows any good online sites that could help me catch up. Here are some of the questions to give you an idea of what I am looking for:

    Question 1: Determine whether the sequence converges or diverges. If it converges, find the
    limit.

    (a) {1 + [(-1)^n]/2 }

    (b) {1 + [(-1)^n]/3n }

    (c) {sin(n)/n}

    Question 2: Determine whether the sequence is increasing, decreasing or not monotonic. Is the sequences bounded? If the sequence is convergent, find its limit.

    (a) { (sqrt(n)) / 1 + sqrt(n) }

    (b) { 2 + 1/3^n }

    Question 3: Determine whether the series is convergent or divergent. If it is convergent, find
    its sum.

    (a) [itex]\sum 4n+2 / 4n - 2[/itex]

    Any advice on some good reference material would be GREATLY appreciated.. Thanks in advance!
     
  2. jcsd
  3. Sep 14, 2011 #2

    lanedance

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  4. Sep 14, 2011 #3

    Char. Limit

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    Re: Convergence/Divergence

    For question three, just see what the summand tends to as n tends to infinity. If it doesn't tend to zero, your sum won't converge.
     
  5. Sep 14, 2011 #4
    Re: Convergence/Divergence

    Wouldn't both the numerator and denominator go to infinity?
     
  6. Sep 14, 2011 #5

    Char. Limit

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    Re: Convergence/Divergence

    Yes, but that doesn't tell us much. For this one, you can rewrite the function as such:

    [tex]\frac{4n+2}{4n-2} = \frac{4n-2+4}{4n-2} = 1 + \frac{4}{4n-2}[/tex]

    Try evaluating the limit from here.
     
  7. Sep 14, 2011 #6
    Re: Convergence/Divergence

    So it is convergent? And it's sum is 1?
     
  8. Sep 14, 2011 #7

    Char. Limit

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    Re: Convergence/Divergence

    No, that's not the series. That's just the summand, the term in the series. You do know the limit test, right?

    Limit test:
    If an does not tend to 0 as n tends to infinity, then [itex]\sum a_n[/itex] diverges.
     
  9. Sep 14, 2011 #8
    Re: Convergence/Divergence

    Like I said, haven't been to class yet. I'll try and get some notes tomorrow, thanks for the help though I appreciate it man.
     
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