Putnam 2021, Problem B-2

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:
The answer is $2/3$.

By AM-GM, we have
$$2^{n+1}(a_1 \cdots a_n)^{1/n} = \left((4a_1)(4^2a_2) \cdots (4^na_n)\right)^{1/n}$$
$$\leq \frac{\sum_{k=1}^n (4^k a_k)}{n}.$$

Thus
$$2S \leq \sum_{n=1}^{\infty} \frac{\sum_{k=1}^n (4^k a_k)}{4^n}$$
$$= \sum_{n=1}^{\infty} \sum_{k=1}^n (4^{k-n} a_k) = \sum_{k=1}^{\infty} \sum_{n=k}^{\infty} (4^{k-n} a_k)$$
$$= \sum_{k=1}^{\infty} \frac{4a_k}{3} = \frac{4}{3}$$
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 $n$.