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Sequence problem

  1. Aug 17, 2013 #1
    1. the nsider, for n → 1, the sequence an given by

    an = n log (n/n+1)

    Determine the limit of the sequence as n→∞, If it exists , or explain why the sequence diverges. In your answers include the names of any rules, theorems or limits you have used.

    2. Relevant equations

    3. The attempt at a solution

    lim n→∞ an = lim n→∞ n log(n/n+1)

    embed sequence in function f(x)= x log (x/(x+1))

    now limx→∞ x log(x/x+1) which is (∞ . ∞) form, can rearange to give

    = limx→∞ log(x/(x+1)) / (1/x) which gives ( 0 / 0 ) indeterminate form, can then use
    log laws on the top function to rewrite log(x/(x+1)) as log(x)-log(x+1)

    = limx→∞ log(x) - log (x+1) / (1/x)

    now differntiate numerator and denominator by l'hopital rule.

    = limx→∞ ((1/x) -(1/(1+x)) / (-1 /x^2) The x(1+x) terms cancel out when you flip and multiply. Leaving

    = limx→∞ 1 x (-1) = -1

    Therefore the limit approaches -1 and converges to -1.
    Using theorem limx→∞ f(x) = L => lim n→∞ an = L.

    So limn→∞ an converges to -1.

    Any help or tips would be appreciated as my Maths is really bad.
     
    Last edited by a moderator: Aug 17, 2013
  2. jcsd
  3. Aug 17, 2013 #2
    You have arrived at the correct answer.
     
  4. Aug 17, 2013 #3

    LCKurtz

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    Science Advisor
    Homework Helper
    Gold Member

    Your work looks OK. Another simpler approach would be to write$$
    a_n = n\log \frac n {n+1} = \log\left( \frac n {n+1}\right)^n =
    \log\left(\frac 1 {1 +\frac 1 n}\right)^n$$Do you see how to finish easily?
     
  5. Aug 17, 2013 #4

    Evo

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    Staff: Mentor

    Please remember to ALWAYS quote the OP's posts.
     
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