Answer
convergent
Work Step by Step
Given
$$ \sum_{n=1}^{\infty} \frac{ 1}{\sqrt[3]{3n^{4}+1}}$$
Use the Limit Comparison Test with $a_{n}=\dfrac { 1}{\sqrt[3]{3n^{4}+1}}$ and $b_{n}=\dfrac { 1}{\sqrt[3]{ n^{4} }}$
\begin{align*}
\lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}&=\lim _{n \rightarrow \infty} \frac{ \sqrt[3]{ n^{4} }}{\sqrt[3]{3n^{4}+1}}\\
&=\lim _{n \rightarrow \infty} \frac{ \sqrt[3]{ n^{4}/n^4 }}{\sqrt[3]{3+1/n^4}}\\
&= \frac{1}{\sqrt[3]{3}}
\end{align*}
since $\displaystyle\sum_{n=1}^{\infty} \frac{ 1}{\sqrt[3]{ n^{4} }}$ is convergent $(p-\text { series } p>1),$ then $\displaystyle\sum_{n=1}^{\infty} \frac{ 1}{\sqrt[3]{3n^{4}+1}}$ also convergent