Answer
$\{a_{n}\}$ diverges to infinity.
Work Step by Step
For any given (large) positive number M, we can find an N such that
$n\gt N\Rightarrow a_{n}\gt M:$
\begin{align*}
N-\displaystyle \frac{1}{N}&\gt M \\
\displaystyle \frac{N^{2}-1}{N}&\gt M \\
N^{2}-1&\gt MN\\
N^{2}-MN-1&\gt 0 \end{align*}
The function $f(x)=x^{2}-Mx-1$
has two distinct zeros as the discriminant $b^{2}-4ac=M^{2}+4$ is positive.
Let $x_{2}$ be the greater of the two zeros.
Its graph is a parabola that opens up, so, for $x\gt x_{2},\ f(x)$ is positive.
We take N to be the first positive integer greater than $x_{2}$ and we are guaranteed that
$ n\gt N\Rightarrow a_{n}\gt M$,
because $n^{2}-Mn-1\gt 0$
Thus, by definition, $\{a_{n}\}$ diverges to infinity.
