$\sum{\frac{(-1)^{n-1}}{{(n(n+1))^{1/3}}}}$ series convergence $$\sum{\frac{(-1)^{n-1}}{{(n(n+1))^{1/3}}}}$$
Does this converge or diverge (absolute and/or conditional)?$\\$
I've tried Leibniz, D'Alembert, Cauchy and Cauchy-integral criteria, all that's left is the comparison test (those are the only things I can use). Any hints would be helpful.
 A: One may use the alternating series test, just check that
$$
n \mapsto \frac{1}{{(n(n+1))^{1/3}}}
$$ is monotonically decreasing over $[1,\infty)$ and observe that, as $n \to \infty$,
$$
\frac{1}{{(n(n+1))^{1/3}}} \to 0.
$$
A: Leibniz's Test is fine here: the series is alternating and $1/(n(n+1))^{1/3}$ decreases to zero. Hence the series is convergent.
A: For absolute convergence:
Let $a_n=\frac1{(n(n+1))^{1/3}}$. Now,
$$\sum\frac1{(n(n+1))^{1/3}}=\sum\frac1{(n^2+n)^{1/3}}=\sum\frac1{n^{2/3}(1+\frac1n)^{1/3}}$$
Let $b_n=\frac1{n^{2/3}}$ (divergent) 
$$\lim \frac{a_n}{b_n}=\lim \frac1{(1+\frac1n)^{1/3}}=1\text{ (finite and non-zero)}$$
By Limit Comparison Test, the series is not absolutely convergent.
For conditional convergence:
It is easy to see that $\frac{1}{(n(n+1))^{1/3}}$ is monotonically decreasing and converges to $0$. So by Leibniz's Test it is convergent.
Note: Leibniz test says that $\sum (-1)^na_n$ converges. But you can write this as $\sum(-1)^{n-1}a_{n-1}$. Therefore, it doesn't make a difference that you have $(-1)^{n-1}$ in your series instead of $(-1)^n.$
