Answer
Converges
Work Step by Step
The Comparison Test states that the p-series $\sum_{n=1}^{\infty}\frac{1}{n^{p}}$ is convergent if $p\gt 1$ and divergent if $p\leq 1$.
Using the Comparison Test , the series is less than $\Sigma _{k=1}^{\infty}\frac{\sqrt[3] k}{\sqrt{ k^{3}}} $ because the denominator of the original series is always slightly larger than the series on the right.
Therefore,
$\Sigma _{k=1}^{\infty}\frac{\sqrt[3] k}{\sqrt{ k^{3}+4k+3}} \leq \Sigma _{k=1}^{\infty}\frac{\sqrt[3] k}{\sqrt{ k^{3}}} $
Re-arranging the exponents .
$\Sigma _{k=1}^{\infty}\frac{\sqrt[3] k}{\sqrt{ k^{3}+4k+3}} \leq \Sigma _{k=1}^{\infty}\frac{k^{1/3}}{k^{3/2}} $
$\Sigma _{k=1}^{\infty}\frac{\sqrt[3] k}{\sqrt{ k^{3}+4k+3}} \leq \Sigma _{k=1}^{\infty}\frac{1}{k^{1/6}} $
We find that the series on the right side is a p-series with $p \gt 1$ , so the series on the right side converges and therefore, the series on the left converges.
