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 $2^{n}$.
Is the answer that simple?
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
