Sum of Products What is the smallest upper bound for the following:
\begin{eqnarray*}
A\left(N\right):=\max_{a_{1}a_{2},...,a_{N}\in\left[0,1\right]}\sum_{n=1}^{N}a_{n}^{1/2}\prod_{s=1}^{n-1}\left(1-a_{s}\right)^{1/2}
\end{eqnarray*}
where empty product is one? I can only get $A\left(N\right)\leq C\ln N$
for some $C<\infty$. For example, let $a_{n}=1/\left(2n\right)^{2}$,
then 
\begin{eqnarray*}
\\1>\prod_{s=1}^{n-1}\left(1-a_{s}\right)^{1/2} & > & \prod_{s=1}^{\infty}\left(1-a_{s}\right)^{1/2}\\
 & = & (2/\pi)^{1/2}
\end{eqnarray*}
using the formula 
\begin{eqnarray*}\sin\left(\pi x\right)=\pi x\prod_{s=1}^{\infty}\left(1-\left(\frac{x}{s}\right)^{2}\right).
\end{eqnarray*}
Then, 
\begin{eqnarray*}
A\left(N\right) & < & \sum_{n=1}^{N}\frac{1}{2n}\\
 & \leq & \ln N.
\end{eqnarray*}
Does any other sequence $a_{1}a_{2},...,a_{N}\in\left[0,1\right]$
exist such that the bound is larger than $O\left(\ln N\right)$?

If $a_n=\Theta(n^{-1-c})$ with $c\gt0$ then $A(N)=\Theta(N^{1/2-c/2})$.
