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Proving the inequality

  1. Sep 27, 2005 #1
    I encountered this problem in one of my math lecture notebooks; Our professor didnt show how it was done, so that leaves me clueless. The problem was to show that the sequence {a_n} defined by
    [tex]
    a_1 = 1,
    a_2 = \int^2_1 \frac{dx}{x},
    a_3 = \frac{1}{2},
    a_4 = \int^3_2 \frac{dx}{x} ,
    ...
    [/tex]

    When generalized gives For any natural number n,

    [tex]
    a_{2n-1} = \frac{1}{n}
    [/tex]
    [tex]
    a_{2n} = \int^{n+1}_n \frac{dx}{x} = \ln x |^{n+1}_{n} = \ln \frac{n+1}{n}
    [/tex]

    is decreasing, that is,
    [tex]
    \frac{1}{n} > \ln \frac{n+1}{n} > \frac{1}{n+1}
    [/tex]

    I've tried math induction but I'm stuck at the (ii) part of math induction, and i tried comparing their derivatives, but I can't conclude anything from doing so. I've tried to compute for their areas, but that got me nowhere. I've graphed their functions using a graphing program, and I saw that it is true, but I would like know how i can prove this without graphing...

    thanx in advance for all help and advice on my problem
     
  2. jcsd
  3. Sep 27, 2005 #2

    lurflurf

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    Homework Helper

    when a<b
    [tex](b-a)\min(f)\leq\int_a^b f dx\leq(b-a)\max(f)[/tex]
     
  4. Sep 27, 2005 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Lovely! Absolutely lovely!
     
  5. Sep 27, 2005 #4
    :surprised

    whoa... that is absolutely lovely... boy do I need a refresh on my calculus... :rofl:

    thanx lurflurf! you're a lifesaver :)
     
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