If $\frac {a_{n+1}}{a_n} \ge 1 -\frac {1}{n} -\frac {1}{n^2}$ then $\sum\limits_na_n$ diverges

You may notice that $$ \frac{a_{n+1}}{a_n} \geq \left(1-\frac{1}{n}\right)\left(1+\frac{1}{n^2-n-1}\right)^{-1} \tag{1}$$ hence: $$ \frac{a_{N+1}}{a_2}\geq \frac{1}{N}\prod_{n=2}^{N}\left(1+\frac{1}{n^2-n-1}\right)^{-1} \tag{2}$$ but the infinite product $\prod_{n\geq 2}\left(1+\frac{1}{n^2-n-1}\right)^{-1}$ is convergent to a positive number ($-\frac{1}{\pi}\cos\frac{\sqrt{5}}{2}\approx 0.296675134743591$). It follows that $a_{N+1}\geq \frac{C}{N}$ so $\sum_{n\geq 2}a_n$ is divergent.

It's easy to show that, for every $n\ge3$, $$ 1 -\frac {1}{n} -\frac {1}{n^2} \ge \frac {n-2}{n-1}$$ It follows that, for every $n\ge3$, $$\frac{a_{n+1}}{a_n}\ge \frac {n-2}{n-1}$$ Thus, $$\frac {a_4}{a_3} \ge \frac 1 2\qquad \frac {a_5}{a_4} \ge \frac 2 3\qquad \ldots\qquad \frac {a_{n-1}}{a_{n-2}} \ge \frac {n-4}{n-3}\qquad \frac {a_n}{a_{n-1}} \ge \frac {n-3}{n-2}$$

Multiplying all the above, one gets: $$\frac {a_n}{a_3} \ge \frac 1 {n-2}$$ hence $$a_n\ge \frac{a_3}{n-2}$$ The last inequality together with the comparison criteria for series proves the divergence.