Answer
Diverges
Work Step by Step
Apply the limit comparison test.
Therefore, $ \lim\limits_{k \to \infty} \dfrac{a_k}{b_k}=\lim\limits_{k \to \infty} \dfrac{k^2/k^3+1}{1/k}\\=\lim\limits_{k \to \infty} \dfrac{k^3}{k^3+1}\\=\lim\limits_{k \to \infty} \dfrac{k^3+1}{k^3+1} -\dfrac{1}{k^3+1} \\=\lim\limits_{k \to \infty} 1- \dfrac{1}{k^3+1} \\=1-0\\=1 \ne 0 \ne \infty$
We can conclude that the given series diverges by the limit comparison test because $\Sigma_{n=1}^{\infty} \dfrac{1}{k}$ is a p-series with $p=1$, so it diverges.
