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√n + 1/n monotone proof

  1. Nov 2, 2012 #1
    Hi,

    Any idea how I may prove that √n + 1/n is monotone? I am unable to show that √(n+1) + 1/(n+1) > √n + 1/n. Thanks!
     
  2. jcsd
  3. Nov 2, 2012 #2

    Mark44

    Staff: Mentor

    Are you required to use induction? That seems to be what you're doing.

    A different approach would be to show that if f(x) = x1/2 + x-1, then f'(x) > 0 if x is large enough (it doesn't have to be very large).
     
  4. Nov 2, 2012 #3

    Ray Vickson

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    The result is true if you mean √(n + (1/n)) and false if you mean what you *wrote*, which means (√n) + (1/n) (although in that case it is true for n ≥ 2).

    RGV
     
  5. Nov 2, 2012 #4
    Regardless of induction, an ascending monotone sequence satisfies the following:
    There exist n0 natural, so that for every n>n0, the n+1th term of the sequence is greater than the nth term.
    And I am not allowed to use functions.
     
  6. Nov 3, 2012 #5

    Mark44

    Staff: Mentor

    Why not? You need to show us the complete problem statement.
     
  7. Nov 3, 2012 #6
    The reason I am not allowed to use functions as we haven't dealt with them in class yet.
     
  8. Nov 3, 2012 #7

    Mark44

    Staff: Mentor

    But surely you have studied functions in a previous class. The function f(x) = √x + 1/x is defined for real x > 0, and gives the same values at positive integers as the sequence sn = s(n) = √n + 1/n. Looking at the graph of f can give you some insight into what your sequence does.
     
  9. Nov 5, 2012 #8
    As I am not permitted to use functions, could someone please suggest how I may prove that the above sequence is monotone?
     
  10. Nov 5, 2012 #9

    uart

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    As Ray has already pointed out, your series is not monotonic. Just calculate the first three terms to demonstrate this.

    Or were you given a more specific problem statement?
     
  11. Nov 5, 2012 #10
    It is monotone, as for any n>=2 a_n+1>a_n
     
  12. Nov 5, 2012 #11

    uart

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    Good, so the question I was asking is whether you were given "for n>=2" as part of the problem statement or if you were supposed to find for what range of "n" it was monotone. In any case, proceed as follows.

    Start by rearranging the reciprocals on one side of the inequation and the square roots on the other. Simplify the reciprocals to a common denominator and then multiply each side by the conjugate surd, [itex](\sqrt{n+1} + \sqrt{n})[/itex]. It's relatively easy after that.
     
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