Does $\sum^\infty_{n=3}\frac {\ln^2(n)}{n^{3/2}}$ converge or diverge? I am trying to find a way to decide if this sum converges or diverges, I'm trying to use the integral test and bound the integrand or use the limit comparison test, but I am not finding a good function to compare my integrand with at infinity.  I would appreciate any help or tips in how to find a function that behaves like $\frac {\ln^2(x)}{x^{3/2}}$ at $x\to \infty$ that I can use it for limit comparison test. 
Thanks in advance.

Integral test for the convergence. The integral is pretty doable
$$\int_3^{\infty } \frac{\log ^2(x)}{x^{3/2}} \, dx=\lim_{M\to\infty}\left[-\frac{2 \left(\log ^2(x)+4 \log (x)+8\right)}{\sqrt{x}}\right]_3^M=\frac{2 \left(8+\log ^2(3)+\log (81)\right)}{\sqrt{3}}$$
As the integral converges, the series converges as well.
