Answer
$$
1+ \frac{ 1}{ \sqrt[3]{ 4}}+\frac{ 1}{ \sqrt[3]{ 9}}+\frac{ 1}{ \sqrt[3]{ 16}}+\frac{ 1}{ \sqrt[3]{ 25}} ...
$$
The given series diverges.
Work Step by Step
$$
1+ \frac{ 1}{ \sqrt[3]{ 4}}+\frac{ 1}{ \sqrt[3]{ 9}}+\frac{ 1}{ \sqrt[3]{ 16}}+\frac{ 1}{ \sqrt[3]{ 25}} ...
$$
this series can be written as
$$
\begin{aligned}
1+ \frac{ 1}{ \sqrt[3]{ 2^ {2}}}+\frac{ 1}{ \sqrt[3]{ 3^ {2}}}+\frac{ 1}{ \sqrt[3]{ 4^ {2}}}+\frac{ 1}{ \sqrt[3]{ 5^ {2}}} ... &=
\sum_{n=1}^{\infty} \frac{1}{\sqrt[3] {n^{2}}}\\
&=\sum_{n=1}^{\infty} \frac{1}{(n^{2})^{ \frac{1}{3}}} \\
&=\sum_{n=1}^{\infty} \frac{1}{n^{\frac{2}{3}}}
\end{aligned}
$$
Apply p-series we obtain $ 0 \lt p=\frac{2}{3 } \lt 1$.
So, the given series is diverges.