Showing $\sum\limits_{n=1}^{\infty} \frac{(-1)^n n^3}{(n^2 + 1)^{4/3}}$ diverges We have the series $\sum\limits_{n=1}^{\infty} \frac{(-1)^n n^3}{(n^2 + 1)^{4/3}}$. I know that it diverges, but I'm having some difficulty showing this. The most intuitive argument is perhaps that the absolute value of the series behaves much like $\frac{n^3}{\left(n^2\right)^{4/3}} = \frac{1}{n^{-1/3}}$, which diverges, though this doesn't seem like it would disprove thee fact that we could be dealing with a conditionally-convergent series. The computation of the limit, even of the absolute value of the general term, also seems nearly impossible to do by hand, as successive applications of L'Hospital's Rule seem to produce a result just as disorderly as what I started with. Limit comparison also doesn't quite seem to work, especially with the alternating-factor.
Thanks in advance for any insights on this. 
 A: Your reasoning is fairly sound, but you are thinking about it a little too hard. Try, instead of thinking about this as a Series and trying to get to a p-series test; do a test for divergence (or nth term test depending on your book) to see if the term itself goes to zero with $n$.
Edit:
Sorry missed the last part of your question, I thought you were trying to compute the series by hand.
Notice that $a_n \rightarrow 0$ iff $|a_n| \rightarrow 0$ for this sequence. A couple applications of L'Hospitals should get you to something that you can determine the convergence/divergence of I would think... (in fact, 1 application and then simplifying, should get you to an end result I believe).
Alternatively, for those that don't like L'Hospital's rule, you can divide the top and bottom by $x^3$. It will help to write the bottom $x^3$ as $(x^9)^{\frac{1}{3}}$ (Edit: Reading is gud. Fixed $x^2$ to $x^3$)
A: Note that
$$ \left|\frac{(-1)^n n^3}{(n^2 + 1)^{4/3}}\right|\sim {n^\frac13}\to \infty$$
thus the given series diverges since each terms in the limit diverges.
Recall indeed for convergence we always need that, as necessary condition, that $|a_n|\to 0$.
