You have
$$
\frac{1}{2^m m^{3/2}} \approx \frac{1}{n},
$$
or equivalently,
$$
2^m m^{3/2} \approx n
$$
and taking logs,
$$
m + \frac{3}{2} \log m \approx \log n
$$
So
$$
\frac{n}{\log n} \approx \frac{2^m m^{3/2}}{m + \frac{3}{2} \log m}
$$
From your final line, $$ \lim_{n\rightarrow\infty}\frac{2^{m(n)}\log n}{n}=\lim_{n\rightarrow\infty}\frac{m+\frac{3}{2}\log m}{m^{3/2}}. $$ Then it converges to 0. So this is not what I want to get.
Discussion between kayak and gt6989b
Discussion between kayak and gt6989b