How can I show that the series [(-1)^n]/n^(1/3) converges?
$\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt[3]{n}}$
By the alternating series test since $\displaystyle \forall{n}>0,\frac{1}{\sqrt[3]{n}}>\frac{1}{\sqrt[3]{n+1}}$
so for an arbitrary value we have showed that for every value greater than that value that
$\displaystyle a_{n+1}<a_n$
Now we must show that $\displaystyle \lim_{n\to\infty}\bigg|\frac{(-1)^n}{\sqrt[3]{n}}\bigg|=\lim_{n\to\infty}\frac{1}{\sqrt[3]{n}}=0$
So we have also established that $\displaystyle \lim_{n\to\infty}|a_n|=0$
So this series conditionally converges
but it does not absolutely converge since
$\displaystyle \sum_{n=1}^{\infty}\bigg|\frac{(-1)^n}{\sqrt[3]{n}}\bigg|=\sum_{n=1}^{\infty}\frac{1}{\sqrt[3]{n}}$
diverges by the integral test
since it meets all its requirements and
$\displaystyle \int_1^{\infty}\frac{1}{\sqrt[3]{n}}dn=\infty$
Assuming you mean $\displaystyle \sum_{n=1}^{\infty}\bigg[\frac{(-1)^n}{\sqrt{n}}\cdot\frac{(-1)^n}{\sqrt[3]{n}}\bigg]$ and not $\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt{n}}\cdot\sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt[3]{n}}$
then yes you are correct the multiplication of both would yield $\displaystyle \frac{1}{n^{\frac{5}{6}}}$
which diverges by the integral test