Answer
Diverge by Th.4.12
Work Step by Step
Th. 9.12 Direct Comparison Test:
Let $0 < a_{n} \leq b_{n}$ for all $n$.
1. If $\displaystyle \sum_{n=1}^{\infty}b_{n}$ converges, then $\displaystyle \sum_{n=1}^{\infty}a_{n}$ converges.
2. If $\displaystyle \sum_{n=1}^{\infty}a_{n}$ diverges, then $\displaystyle \sum_{n=1}^{\infty}b_{n}$ diverges.
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$\displaystyle \sum_{n=1}^{\infty}\frac{1}{\sqrt[4]{n}}$ is a divergent p-series $(p=\displaystyle \frac{1}{4} < 1 )$ so we aim for case (2).
Letting $a_{n}=\displaystyle \frac{1}{4\sqrt[4]{n}}$,, we have
$b_{n}=\displaystyle \frac{1}{4\sqrt[3]{n}-1} > \frac{1}{4\sqrt[4]{n}} > 0,$
$0 < a_{n} \leq b_{n}$, and
$\displaystyle \sum_{n=1}^{\infty}a_{n}=\frac{1}{4}\sum_{n=1}^{\infty}\frac{1}{\sqrt[4]{n}}$ diverges , so
$\displaystyle \sum_{n=1}^{\infty}b_{n}=\sum_{n=1}^{\infty}\frac{1}{4\sqrt[3]{n}-1}$ also diverges.