11. (15 points) The sequence of positive integers $\left\{a_{n}\right\}$ satisfies:
$$
a_{4}=16,-\frac{1}{a_{n+1}^{2}}<\frac{1}{a_{n}}+\frac{1}{a_{n+1}}-\frac{2}{n(n+1)}<\frac{1}{a_{n}^{2}} \text {. }
$$

Find the general term formula $a_{n}$.
11. According to the condition $\frac{1}{a_{n+1}^{2}}\frac{1}{16^{2}}+\frac{1}{6}-\frac{1}{16}=\frac{83}{768}$.
Then $a_{3}0$, and
$a_{3} \geqslant 3$, so
$a_{3}>\frac{48+8 \sqrt{21}}{10}>\frac{48+8 \times 4.5}{10}=8.4$. (3)
According to equations (2) and (3), we get the integer $a_{3}=9$.
Again, taking $n=2$ in the condition, we get
$\frac{1}{a_{3}^{2}}\frac{1}{81}+\frac{2}{9}=\frac{19}{81}$.
Then $a_{2}0$, i.e.,
$\left(a_{2}-3\right)\left(2 a_{2}-3\right)>0$.
Since $a_{2} \geqslant 2$, therefore, $a_{2}>3$. Hence $a_{2}=4$. Then, taking $n=1$ in the known equation, we get
$\frac{1}{16}\frac{13}{16}$, so $a_{1}\frac{n^{3}+n^{4}}{n^{3}-n^{2}+n+1}$
$=(n+1)^{2}-\frac{2 n^{2}+3 n+1}{n^{3}-n^{2}+n+1}$
$>(n+1)^{2}-1$.
(Here we use the fact that when $n \geqslant 4$,
$\left(n^{3}-n^{2}+n+1\right)-\left(2 n^{2}+3 n+1\right)$
$=n\left(n^{2}-3 n-2\right)=n[n(n-3)-2]$
$\geqslant n(n-2)>0$.)
According to the left side of equation (7), we get $\frac{1}{a_{n+1}^{2}}0$.
Assume that the two roots of the quadratic equation corresponding to equation (9) are $\alpha, \beta(\alpha0$.
According to equations (8) and (9), we get $a_{n+1}=(n+1)^{2}$.
Therefore, by induction, for any $n \in \dot{\mathbf{N}}$, equation (6) holds.