Did I solve this problem correctly? $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{n^2}{n^3+4}$ For the following series:
$\sum_{n=1}^{\infty}(-1)^{n+1}\frac{n^2}{n^3+4}$
I found $b_n$ to be $b_n=\frac{n^2}{n^3+4} \gt 0$ and then I took the limit of this and found it to be zero.

I propose another way to the result. Let $u_n = \dfrac{(-1)^{n+1} n^2}{n^3+4}$ and $v_n = \dfrac{(-1)^{n+1}}{n}$.
We already know that the series $\sum v_n$ is convergent (by the alternating test for example). Now, consider the difference $|u_n-v_n|$. We have
$$
\sum_{n=1}^\infty \left|u_n - v_n\right| = \sum_{n=1}^\infty \frac{1}{n} \frac{\frac{4}{n}}{1+\frac{4}{n}} \leq \sum_{n=1}^\infty \frac{4}{n^2} <\infty,
$$
a consequence of which is that $\sum u_n = \sum v_n + \sum (u_n - v_n)$ is also convergent because it is a sum of two convergent series.
Remark. This method is particularly useful when the alternating test fails.
