Answer
Converges
Work Step by Step
Apply the integral test. Let us consider that $\Sigma a_k$ is a series with positive terms such that $a_k=f(k)$ for all $k \geq a$:
$\Sigma_{k=1}^{\infty}$ and $\int_a^{\infty} f(x) \ dx$ diverge or converge.
We can see that $\Sigma_{k=1}^{\infty} \dfrac{\sqrt k \ln k}{k^3+1} \lt \dfrac{k \ln k}{k^3}=\dfrac{\ln k}{k^2}$
Therefore, $\int_2^{\infty} \dfrac{\ln x}{x^2} \ dx=\lim\limits_{z \to \infty}\int_2^{z} \dfrac{\ln x}{x^2} \ dx\\=\lim\limits_{b \to \infty} [\dfrac{-\ln x}{x}-\dfrac{1}{x}]_{2}^z\\=\dfrac{1}{2}[\ln (2)+1] \lt \infty$
So, we can conclude that the given series converges by the integral test.
