$a_{3n} = {-1}/ \sqrt[3]n, \ a_{3n+1} = {-2}/ \sqrt[3]n, \ a_{3n+2}= {3} / \sqrt[3]n$ then $\sum_{n=1}^{\infty} a_n$ converges Let $a_n$ defined by:
$$a_{3n} = \frac{-1}{\sqrt[3]n},\quad a_{3n+1} = \frac{-2}{\sqrt[3]n},\quad a_{3n+2}=\frac{3}{\sqrt[3]n} $$
show that $\sum_{n=1}^{\infty} a_n$ converges.
I thought about looking at partial sums, but I don't see how they converge...

Let $(c_n)$ be the sequence given by $-2, 3,-1, -2, 3, -1, -2, 3, -1, \cdots$.
Then $(c_n)$ has bounded partial sums, and the sequence $\left(\frac{1}{\sqrt[3]{n}}\right)$ is decreasing to 0,
so the series $\sum\frac{c_n}{\sqrt[3]{n}}$ converges by Dirichlet's test.
