We have $\sum_{n=0}^\infty (-1)^n \frac{n^2-1}{n^2}$ with $a_n=(-1)^n \frac{n^2-1}{n^2}$
$i$) Notice that $\lim_{n\to\infty}|a_n|=1\not=0$. Therefore, $\sum_{n=1}^\infty |a_n|$ diverges and the series given are not absolutely convergent.
$ii$) Let $a_{2n}$ and $a_{2n+1}$ be two sub-sequences of $a_n$ of its positive and negative terms respectively. Notice that $\lim_{n\to\infty}\frac{2n-1}{2n+1}=1 \not= \lim_{n\to\infty} -\frac{2n}{2n+2}=-1$. Because $a_n$ has two sub-sequences that converge to different limits, $a_n$ diverges; therefore $\lim_{n\to\infty}a_n \not=0$ and this implies that $\sum_{n=0}^\infty a_n$ diverges.
That is correct, but it is much simpler to note that, since you dont have $\lim_{n\to\infty}(-1)^n\frac{n^2-1}{n^2}=0$, the series diverges.