Polynomial-Exponential Equation

Question

Consider the sequence $$ x_n = \sup \{ k \in \mathbb{N} : e^{ 2^{k}} 2^k \leq n \} $$ I'm wondering if it possible to deduce that there exists $\alpha \in (0,1)$ for which $ e^{2^{x_n}} = O(n^{\alpha}) $ or a proof as to why no such $\alpha$ exists.

Answer 1

Let $f(x) := 2^x \exp(2^x)$ and note that $$f(m+1) -f(m) \geq (2^{m+1} -2^m) \exp(2^m) \geq 2.$$ Thus $f(m) \leq n < f(m+1)$ with $n := \lceil f(m) \rceil$ and therefore $x_n = m$. So $$\exp(2^{x_n}) = \exp(2^m) = f(m)/2^m \geq \frac{n-1}{2^m}.$$ Since $n \geq \exp(2^m)$, we also have $2^m \leq \ln(n)$ and thus $$\exp(2^{x_n}) \geq \frac{n-1}{\ln(n)}.$$ We conclude that no $\alpha \in (0,1)$ exists such that $\exp(2^{x_n}) = O(n^\alpha)$.