Sequence queue - converges or diverges? I have the next sequence queue: $\sum_{n=1}^{\infty}{\frac{n^2-n-1}{n^4+n^2+1}}$. Does the queue converges or diverge?

My attempt:
I have tried to show that $\frac{1}{n^2}>|\frac{n^2-n-1}{n^4+n^2+1}|$ for any $n>0$, and then by the comparation test we get that since $\sum_{n=1}^{\infty}{\frac{1}{n^2}}$ converges, we have that $\sum_{n=1}^{\infty}{|\frac{n^2-n-1}{n^4+n^2+1}|}$ converges too, and by a theorem we have that since $\sum_{n=1}^{\infty}{|\frac{n^2-n-1}{n^4+n^2+1}|}$ $\implies$ $\sum_{n=1}^{\infty}{\frac{n^2-n-1}{n^4+n^2+1}}$ converges. Is that right?

We are trying to prove that the series, $$\sum_{n=1}^{\infty}{\frac{n^2-n-1}{n^4+n^2+1}}$$
converges.
Let us use the Limit Comparison Test. That is, if $a_n>0$ and $b_n>0$, then if $$\lim_{n\to \infty}\frac{a_n}{b_n}=L$$ for some real number $L$. Then, if $b_n$ converges, by the limit comparison test, $a_n$ must also converge.
Let $a_n=\frac{n^2-n-1}{n^4+n^2+1}$ and we say $b_n=\frac{1}{n^2}$. Then, $$\lim_{n\to \infty} \frac{\frac{n^2-n-1}{n^4+n^2+1}}{\frac{1}{n^2}}\\
\lim_{n\to \infty}\left(\frac{(n^2-n-1)(n^2)}{n^4+n^2+1}\right)\\
\lim_{n\to \infty}\frac{n^4-n^3-n^2}{n^4+n^2+1}\cdot \frac{\frac{1}{n^4}}{\frac{1}{n^4}}\\
\lim_{n\to \infty} \left(\frac{{1-\frac{1}{n}}-\frac{1}{n^2}}{1+\frac{1}{n^2}+\frac{1}{n^4}}\right)\\
\lim_{n\to \infty}\frac{1-0-0}{1+0+0}=\frac{1}{1}=1$$
Hence, we have shown that the $\lim_{n\to \infty}\frac{a_n}{b_n}=L$ converges because it equals some constant; in this case, $1$.
Now, because $\sum_{n=1}^{\infty}\frac{1}{n^2}$ converges by the p-test, (since $p=2>1$), we can conclude that the series, $$\sum_{n=1}^{\infty}{\frac{n^2-n-1}{n^4+n^2+1}}$$ must also converge by the limit comparison test.
