## Calculus (3rd Edition)

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.