1. ## finding maximal value

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

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

2. ## Interesting...

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

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

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

I am trying to imagine an induction approach to repeat this last trick on $\displaystyle \frac{a_{2007}}{a_{2008}}$ and so on, but as of right now, the highest value I can generate is $\displaystyle 1004.9999999999...$

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

Now imagine letting $\displaystyle a_{2008}$ be an arbitrarily large number L, and denote $\displaystyle c=\sum_{i=1}^{2007}a_i$. Now the maximal value of $\displaystyle \frac{a_{2008}}{a_{2009}}$ would be gotten by defining $\displaystyle a_{2009}=L+c$, giving it a value of $\displaystyle \frac{L}{L+c}$ which tends to 1 as L grows large. Thus $\displaystyle S<1+\frac12+\frac12+...+\frac12+1=1005$
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 $\displaystyle a_{2008} = L$ then it affects the term $\displaystyle \frac{a_{2007}}{a_{2008}}$, changing the last two terms of the sum from $\displaystyle \frac12+\frac12$ to $\displaystyle \frac{2^{2005}}L + \frac{L}{L+c},$ which tends to 1 as L→∞. So nothing is gained.

I suspect that the value 1004.5 obtained by taking $\displaystyle a_i=2^{i-2}$ for $\displaystyle i>2$ 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:

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

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

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