Answer
Diverges
Work Step by Step
We have: $a_k=\dfrac{\sqrt[3] {k^2+1}}{\sqrt {k^3+2}}$ and $b_k=\dfrac{1}{k}$
Now, we need to apply the limit comparison test.
$L=\lim\limits_{k \to \infty}\dfrac{a_k}{b_k}\\=\lim\limits_{k \to \infty}\dfrac{\dfrac{\sqrt[3] {k^2+1}}{\sqrt {k^3+2}}}{1/k}\\=\lim\limits_{k \to \infty}\dfrac{\sqrt[6] {(k^5+k^3)^2}}{\sqrt [6]{(k^3+2)^3}} \\=\infty$
Therefore, the series diverges by the limit comparison test.
