# Prove that the following series converges using comparsion test

I have the following series: $$\sum_{n= 1}^{\infty}\frac{n^3}{2^{\ln^2n}}$$ and I want to prove that this series converges using comparison test. I already tried with $$b_n := 1/2^n$$ and $$c_n := 1/e^n$$ but it didn't work. Which series can I try ?

$$2^{\ln^2n}\ge n^5$$
that is for $$n\ge n_0$$ such that $$\ln^2 n \ge 5\cdot \log_2 n \implies n\ge e^{\frac 5 {\log 2}}\approx1357.6$$
$$\sum_{n= n_0}^{\infty}\frac{n^3}{2^{\ln^2n}}\le \sum_{n= n_0}^{\infty}\frac{n^3}{n^5}=\sum_{n= n_0}^{\infty}\frac{1}{n^2}$$