Putnam

The answer is $2/3$.

By AM-GM, we have
$$\begin{align*} 2^{n+1}(a_1\cdots a_n)^{1/n} &= \left((4a_1)(4^2a_2)\cdots (4^na_n)\right)^{1/n}\\ & \leq \frac{\sum_{k=1}^n (4^k a_k)}{n}. \end{align*}$$
Thus
$$\begin{align*} 2S &\leq \sum_{n=1}^\infty \frac{\sum_{k=1}^n (4^k a_k)}{4^n} \\ &= \sum_{n=1}^\infty \sum_{k=1}^n (4^{k-n}a_k) = \sum_{k=1}^\infty \sum_{n=k}^\infty (4^{k-n}a_k) \\ &= \sum_{k=1}^\infty \frac{4a_k}{3} = \frac{4}{3} \end{align*}$$
and $S \leq 2/3$. Equality is achieved when $a_k=\frac{3}{4^k}$ for all $k$, since in this case $4a_1=4^2a_2=\cdots=4^na_n$ for all $n$.