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