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**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 $\displaystyle S<1+\frac12+\frac12+...+\frac12+1=1005$