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Basic sequence related question

  1. Jan 14, 2009 #1
    1. The problem statement, all variables and given/known data

    Here is the problem as deployed in the exercise book:
    The point is to say if the sequence converges or diverges
    an= { (-1)^n*(n+1/n) }

    2. Relevant equations

    Bluntly enough: what's the key to finding the result?

    3. The attempt at a solution

    I tried applying l'Hopital's rule but got lost into oblivion. I tried finding a smaller and bigger sequences (sandwich propriety), couldnt acheive it...

    The answer guide says the sequence diverges. Why, and how do you get there?

  2. jcsd
  3. Jan 14, 2009 #2
    Do you mean [tex] (-1)^n * (\frac{n+1}{n}) [/tex] or [tex] (-1)^\frac{n(n+1)}n}[/tex]
  4. Jan 14, 2009 #3
    The first one is the correct sequence. Thanks for cleaning it up :) (sorry, noob alert :/)
  5. Jan 14, 2009 #4
    Double post, see below
  6. Jan 14, 2009 #5
    For even n we have [tex]a_n = 1*\frac{n+1}{n} = \frac{n+1}{n}[/tex]. For odd n we have [tex]a_n = -1*\frac{n+1}{n} = \frac{-(n+1)}{n}[/tex].

    Remember that a sequence converges if and only if there is a positive integer N such that when integers n,m > N, we have [tex] |a_n - a_m| < \epsilon [/tex] for all epsilon greater than 0.

    That should help you out :).
    Last edited: Jan 14, 2009
  7. Jan 14, 2009 #6


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    If your sequence is ((-1)^n)*((n+1)/n) then that's ((-1)^n)*(1+1/n). The limit of the second factor is 1. So for large n your sequence looks like +1,-1,+1,-1,... Is that convergent?
  8. Jan 15, 2009 #7


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    Divide both numerator and denominator by n:
    [tex](-1)^n\frac{1+\frac{1}{n}}{1}= (-1)^n\left(1+ \frac{1}{n}\right)[/tex]
    If this were just
    [tex]\left(1+ \frac{1}{n}\right)[/tex]
    then it should be easy to see what the limit is as n goes to infinity. But with the [itex](-1)^n[/itex] think about the subsequences for n even or odd.
  9. Jan 15, 2009 #8
    Well, that's much more simpler that way! I havent done calculus in 3 years so I kinda lost my intuition. Ugh! Thank you very much people!
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