1. ## 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 $2 \leq \imath \leq 2009$

2. ## 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 $1004.9999999999...$

3. Originally Posted by Media_Man
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.

4. i havent understood a single word sorryy!!!!

5. ## 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.