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Equivalent of 1/n^(a) in the neighborhood of +infinity

  1. Oct 10, 2009 #1
    1. The problem statement, all variables and given/known data
    How can one find an equivalent of 1/n^(a) in the neighborhood of +infinity?
    And what's the method in general to get an asymptotical development of a sequence?

    2. Relevant equations
    The result is:
    1/n equivalent to ln(n+1)-ln(n)
    1/n^a equivalent to (1/a-1)*(1/n^(a-1)-1/(n+1)^(a-1) if a is different from 1.

    3. The attempt at a solution
    I tried to use Taylor Young with f:=x->x in the neighborhood of 0 (this is useless)

    Thanks a lot for your help.
  2. jcsd
  3. Oct 10, 2009 #2


    Staff: Mentor

    No one has responded to your question all day, which might mean that they are as confused about what you are asking as I am. What do you mean "an equivalent of 1/na in the neighborhood of +infinity"?
  4. Oct 11, 2009 #3
    I mean an asymptotical expansion: for instance ln(n+1)-ln(n)=ln(1+(1/n))=(1/n)+o(1/n) (because 1/n is about 0 when n tends to infinity). Therefore 1/n is equivalent to ln(n+1)-ln(n) in the neighborhood of infinity.
    My question is how do we get this equivalent?

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