How to find $b_n$ for the limit comparison test in $\sum_{n=1}^{\infty}\frac{(\ln(n))^2}{\sqrt{n}(10n-9\sqrt{n})}$.

Question

I'm supposed to determine whether or not the series converges or diverges but I'm stuck trying to find $b_n$ and prove that $a_n \sim b_n$. I would very much appreciate it if someone could help show me how I would go about finding $b_n$ proving that $a_n \sim b_n$.

$$\sum_{n=1}^{\infty}\frac{(\ln(n))^2}{\sqrt{n}(10n-9\sqrt{n})}$$

$$a_n=\frac{(\ln(n))^2}{\sqrt{n}(10n-9\sqrt{n})}$$

$$b_n = ?$$

Answer 1

Note that $$a_n\leq \frac{(\ln n)^2}{n^{3/2}}$$ for $n\geq 1$.

You might find the answers here Prove the convergence of : $\sum \ln(n)/n^{3/2}$ helpful for dealing with $\frac{(\ln n)^2}{n^{3/2}}$.