First I introduce a notation similar to $\sum_{i=1}^n a_i$ for exponentiation. I.e. for any (potentially infinite) sequence $a_i$ we define $$ ES_{i=1}^n a_i = \left\{\begin{matrix} a_1 & \text{if }n=1 \\ \left(E_{i=1}^{n-1} a_i\right)^{a_n} & \text{else} \\ \end{matrix}\right.$$
and alternatively
$$ EL_{i=1}^n a_i = \left\{\begin{matrix} a_1 & \text{if }n=1 \\ a_1^\left(E_{i=2}^{n} a_i\right) & \text{else} \\ \end{matrix}\right.$$
We need two such notations because exponentiation is not associative.
Now I want to find the sequences $a_i$ with $\sum_i a_i =1$ maximizing the following expressions
$$ ES_{i=1}^n (1+a_i) \quad\text{and}\quad EL_{i=1}^n (1+a_i)$$
I first thought that taking $a_i = \frac{1}{n}$ will give the solution (similar to how $\Pi_{i=1}^n (1+a_i)$ is maximal when $a_i =\frac{1}{n}$ yielding $e$). And for small $n$ this is indeed increasing, however in the limit case this is not true anymore:
First for $ES$:
\begin{align}\lim_{n\rightarrow\infty}ES_{i=1}^n 1+\frac{1}{n}&=\lim_{n \rightarrow \infty} (1+\frac{1}{n})^{(1+\frac{1}{n})^{n-1}}\\ &=\lim_{n\rightarrow \infty} (1+\frac{1}{n})^{\lim_{n\rightarrow\infty} (1+\frac{1}{n})^{n-1}}\\ &=1^e =1 \end{align}
Then for $EL$ we have \begin{align} EL_{i=1}^n(1+\frac{1}{n})\leq EL_ {i=1}^\infty (1+\frac{1}{n})\end{align}
$EL_{i=1}^\infty x$ is continuous in $x$ (see wikipedia) and therefore this converges to 1.
Hence which sequence maximizes (a maximum will probably not exist, but will converge to the supremum) $EL$ and $ES$?