Prove or disprove: $\lim_{n\rightarrow \infty} \frac{2^n}{a_n} = 0$.

I am trying to determine whether the following statement is true or false:

Let $$a_n$$ be an unbounded non decreasing sequences s.t. $$\sum \frac{1}{\log(a_n)}$$ converges.

Prove or disprove:

$$\lim_{n\rightarrow \infty} \frac{2^n}{a_n} = 0$$.

Since $$\sum \frac{1}{\log(a_n)}$$ converges, $$\log(a_n)$$ must be asymptotic greater than $$n$$, so $$a_n$$ must be asymptotic greater than $$2^{n}$$.

As the sequence $$(\log a_n)^{-1}$$ is nonicreasing the convergence of the series implies (see) $$\lim_n {n\over \log a_n}=0$$ Therefore $${n\over \log a_n}<{1\over 2},\quad n>N$$ i.e. $$a_n>e^{2n},\quad n>N$$ and the conclusion follows.
Remark The assumption that $$a_n$$ is nondecreasing is essential. For example let $$a_n=\begin{cases} e^{n^2} & n\neq 2^{k^2}\\ n & n=2^{k^2} \end{cases}$$ Then $$\sum (\log a_n)^{-1}<\infty$$ but for $$n=2^{k^2}$$ we have $${2^n\over a_n}={2^{2^{k^2}}\over a_{2^{k^2}}}={2^{2^{k^2}}\over 2^{k^2}}\to \infty$$
Let $$a_n=e^{n^2}$$ . $$\sum\frac{1}{log({a_n})}=\sum\frac{1}{n^2}<\infty$$, but $$\lim_{n\rightarrow \infty}\frac{a_n}{2^n}=\infty$$. So it's not true
• There was a typo, $a_n$ should be bellow $2^n$. My bad. Commented Jun 20, 2022 at 17:49