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Proving inequalities with logarithm

  1. Jan 26, 2016 #1
    I need to prove:

    (n+1)*(log(n+1)-log(n) > 1 for all n > 0.

    I have tried exponentiating it and I got

    ( (n+1)/n )^(n+1) < e.

    And from there I couldn't go any farther, but I do know that it is true by just looking at its graph.

    Could anybody help me please?
     
  2. jcsd
  3. Jan 26, 2016 #2

    blue_leaf77

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    Correction:
    As for the solution itself, first find out whether the function
    $$
    f(n) = \left( \frac{n+1}{n} \right)^{(n+1)}
    $$
    is a monotonically decreasing or increasing function for positive ##n## (it should be either of them). Then find the asymptote of this function as ##n \rightarrow \infty## in terms of ##e## by making use of the definition of ##e##
    $$
    e = \lim_{n \to \infty} \left( 1+\frac{1}{n} \right)^n .
    $$
     
  4. Jan 26, 2016 #3
    I tried that, but to know if it is indeed monotonic I have to show that the selected expression in the term the picture I attached is always < 1 or > 1.

    So to proof of the original problem requires a proof for:
    n*(log(n+1)-log(n)) < 1

    Right now it looks pretty circular.
     

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  5. Jan 26, 2016 #4

    mfb

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    You don't need that, there is a more direct way.

    You can also show the initial inequality directly with a clever integral. Can you rewrite ln(n+1)-ln(n) somehow?
     
  6. Jan 26, 2016 #5
    Can I have more clues please? I'm getting nowhere plus I'm not that good with integral calculus since we were never taught this in my high school & university.
     
  7. Jan 26, 2016 #6

    mfb

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    What is ##\int_a^b \frac{dx}{x}##?
     
  8. Jan 26, 2016 #7
    log(x), but I still really can't where I'm supposed to be headed.
     
  9. Jan 26, 2016 #8

    mfb

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    It is not log(x). The definite integral not depend on x at all, but only on a and b.
     
  10. Jan 26, 2016 #9
    log(b)-log(a) ?
     
  11. Jan 26, 2016 #10

    mfb

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    Right, the difference between two logarithms - that's what you have in your original inequality.
     
  12. Jan 26, 2016 #11
    I think you're implying that I integrate both sides, but how do I choose the value of a and b?
     
  13. Jan 26, 2016 #12

    mfb

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    No, don't integrate. Replace (log(n+1)-log(n)) with an integral, and then find some way to get the inequality you need.
     
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