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 $ES$?

  • $\begingroup$ Maybe I'm misunderstanding the question, but aren't you basically looking for the extremum of function $f$ in $n$ variables (by taking a logarithm), $$ f=(1+a_n)(1+a_{n-1})\cdots (1+a_2)\log (1+a_1) $$ conditioned on $\sum_{i=1}^n a_i=1$? Can't you calculate this using a Lagrange multiplier method? $\endgroup$ – Hamed Oct 19 '18 at 6:35
  • $\begingroup$ @Hamed Yes, I guess that's a good approach for $ES$ not however for $EL$. However if I remember Lagrange multipliers method correctly it would only solve for $a_i$ given the $n$. It doesn't give you the optimal $n$, right? $\endgroup$ – Jürg Merlin Spaak Oct 19 '18 at 7:04

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