If $\frac {a_{n+1}}{a_n} \ge 1 -\frac {1}{n} -\frac {1}{n^2}$ then $\sum\limits_na_n$ diverges 
Let $(a_n)_{n \ge 1}$ be a sequence of positive real numbers such that, for every $n\ge1$,
  $$\frac {a_{n+1}}{a_n} \ge 1 -\frac {1}{n}  -\frac {1}{n^2} \tag 2$$ Prove that $x_n=a_1 + a_2 + .. + a_n$ diverges.

It is clear that $x_n$ is increasing, so it has to have a limit. I tried to prove the limit is $+\infty$ but without success. No divergence criteria from series seems to work here.
UPDATE
Attempt: 
Suppose a stronger inequality holds, namely that, for every $n\ge1$, $$\frac{a_{n+1}}{a_n}\geqslant1-\frac1n \tag 1$$ Then:
$$\frac {a_3}{a_2} \ge \frac 1 2\qquad
\frac {a_4}{a_3} \ge \frac 2 3\qquad
\ldots\qquad
\frac {a_{n-1}}{a_{n-2}} \ge \frac {n-3}{n-2}\qquad
\frac {a_n}{a_{n-1}} \ge \frac {n-2}{n-1}$$
Multiplying all the above yields
$$\frac {a_n}{a_2} \ge \frac 1 {n-1}$$
The last inequality proves the divergence.

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.
