Answer
Diverges
Work Step by Step
Given $$\sum_{n=1}^{\infty} n^{-1/3}$$
Since $f(n)= n^{-1/3}$ , is positive, decreasing and continuous for $n\geq 1$, then by using the integral test
\begin{aligned} \int_{1}^{\infty} f(n) dn &=\lim _{t \rightarrow \infty} \int_{1}^{t} n^{-1/3}d n \\
&=\left.\lim _{t \rightarrow \infty}\frac{3}{2} n^{2/3}\right|_{1} ^{t} \\
&=\frac{3}{2}\lim _{t \rightarrow \infty} t^{2/3}-1 \\ &=\infty\end{aligned}
Thus the series diverges.
