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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 ?

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We can use that eventually

$$2^{\ln^2n}\ge n^5$$

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

then

$$\sum_{n= n_0}^{\infty}\frac{n^3}{2^{\ln^2n}}\le \sum_{n= n_0}^{\infty}\frac{n^3}{n^5}=\sum_{n= n_0}^{\infty}\frac{1}{n^2}$$

Refer also to the related

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