Putnam 2021 
Topic: Algebra Analysis Calculus

Problem: Determine the maximum value of the sum \[ S = \sum_{n=1}^\infty \frac{n}{2^n} (a_1 a_2 \cdots a_n)^{1/n} \] over all sequences $a_1, a_2, a_3, \cdots$ of nonnegative real numbers satisfying \[ \sum_{k=1}^\infty a_k = 1. \]

Solution: 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 $n$. The answer is $\boxed{2/3}$.