$$a_n=\frac{1}{n}\left[n\beta\right]+n^2\beta^{n}$$where $0\lt\beta\lt1$
Now since $[n\beta]=n\beta- \{n\beta \}$, we have $$a_n=\beta-\frac{1}{n}\{n\beta\}+n^2\beta^{n}$$
$$\implies a_n-\beta=n^2\beta^{n}-\frac{1}{n}\{n\beta\}\lt n^2\beta^{n}$$. All I need to do now is show $n^2\beta^{n}\lt \epsilon.$
This is where I am getting stuck.
if $|\beta| <1$ then you should compare $\beta^n$ with $n^2$. just look at $\lim_{n \to \infty} \frac{\beta^{-n}}{n^2}$. Then just use l'Hopital's rule (probably twice). This should teach you that $\lim_{n \to \infty} \frac{\beta^{-n}}{n^2}=\infty$, therefore $\beta^n n^2$ will converge to $0$ if $n \to\infty$. The philosophy behind examining this fraction is finding out which one is 'faster' the polynomial expression $(n^2)$ or the exponential expression $(\beta^n)$.
