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Math Help - finding maximal value

  1. #1
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    finding maximal value

    Find the maximal value of the expression
    \frac{a_1}{a_2} + \frac{a_2}{a_3}+...+ \frac{a_{2008}}{a_{2009}}

    where
    a_1,a_2,...,a_{2009} are positive numbers satisfying a_\imath \geq a_1+a_2+...+a_{\imath - 1}   for each  2 \leq \imath \leq 2009
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  2. #2
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    Interesting...

    Well, here's something just to get the ball rolling...

    If you use powers of 2, that is a_i=2^{i-1}, that gives you S=\frac12+\frac12+...+\frac12 = 2008\frac12=1004. Because of the inequality, you can let a_1=a_2=1 and a_i=2^{i-2} for i>2. This gives S=1+\frac12+\frac12+...+\frac12=1004.5

    Now imagine letting a_{2008} be an arbitrarily large number L, and denote c=\sum_{i=1}^{2007}a_i. Now the maximal value of \frac{a_{2008}}{a_{2009}} would be gotten by defining a_{2009}=L+c, giving it a value of \frac{L}{L+c} which tends to 1 as L grows large. Thus S<1+\frac12+\frac12+...+\frac12+1=1005

    I am trying to imagine an induction approach to repeat this last trick on \frac{a_{2007}}{a_{2008}} and so on, but as of right now, the highest value I can generate is 1004.9999999999...
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  3. #3
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    Quote Originally Posted by Media_Man View Post
    If you use powers of 2, that is a_i=2^{i-1}, that gives you S=\frac12+\frac12+...+\frac12 = 2008\frac12=1004. Because of the inequality, you can let a_1=a_2=1 and a_i=2^{i-2} for i>2. This gives S=1+\frac12+\frac12+...+\frac12=1004.5

    Now imagine letting a_{2008} be an arbitrarily large number L, and denote c=\sum_{i=1}^{2007}a_i. Now the maximal value of \frac{a_{2008}}{a_{2009}} would be gotten by defining a_{2009}=L+c, giving it a value of \frac{L}{L+c} which tends to 1 as L grows large. Thus S<1+\frac12+\frac12+...+\frac12+1=1005
    I don't believe that the construction with L and L+c leads to any improvement on the previous total of 1004.5. In fact, if you choose a_{2008} = L then it affects the term \frac{a_{2007}}{a_{2008}}, changing the last two terms of the sum from \frac12+\frac12 to \frac{2^{2005}}L + \frac{L}{L+c}, which tends to 1 as L→∞. So nothing is gained.

    I suspect that the value 1004.5 obtained by taking a_i=2^{i-2} for i>2 is the best possible.
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  4. #4
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    i havent understood a single word sorryy!!!!
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  5. #5
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    Clarity

    Sorry... So far the most efficient selection of a's we have been able to find is the following:

    a_1=1
    a_2=1
    a_3=1+1=2
    a_4=1+1+2=4
    a_5=1+1+2+4=8
    a_6=1+1+2+4+8=16
    ...
    And so on, amounting to a_i=2^{i-2}

    This gives your sum S=\frac{a_1}{a_2}+\frac{a_2}{a_3}+... +\frac{a_{2008}}{a_{2009}} = \frac{1}{1}+\frac{1}{2}+\frac{2}{4}+... +\frac{2^{2008}}{2^{2009}} = 1+\frac12+\frac12+...+\frac12=1004.5

    This is probably the maximum, as Opalg corrected my faulty logic in my last post. What we lack though, is a proof.
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