We will first prove that \(a_n>0\) for all \(n\in\mathbb{N}\).
For \(n=1\), \(a_{1}=\alpha>0\), supposing that \(a_n>0\) for all \(n\in\mathbb{N}\), \(a_{n+1}=a_{n}{\alpha}^{a_{n}}>0\) which proves the induction.
If
\(\bullet\) \(\alpha=1\), then \(a_n=c\) for all \(n\in\mathbb{N}\) and some constant \(c\in\mathbb{R}\), so \(\displaystyle\sum_{n=1}^{\infty}\frac{a_{n}}{n^{\beta}}=c\sum_{n=1}^{\infty}\frac{1}{n^{\beta}}\) which is convergent for \(\beta>1\).
\(\bullet\) \(\alpha\in(0,1)\), then the sequence \(a_{n}\) is stricly decreasing since \(\displaystyle \log\left(\frac{a_{n+1}}{a_{n}}\right)=a_{n}\cdot \log(\alpha)<0\Rightarrow a_{n+1}<a_{n}\) and by the D'Alembert convergence criterion we get
\[\overline{\lim_{n\to\infty}}\left|\frac{\frac{a_{n+1}}{(n+1)^{\beta}}}{\frac{a_{n}}{n^{\beta}}}\right|=\overline{\lim_{n\to\infty}}\left|\frac{a_{n+1}}{a_{n}}\left(1+\frac{1}{n}\right)^{\beta}\right|<1\]
and so the series is convergent for all \(\beta\).
\(\bullet\) \(\alpha>1\), then \(a_n\) is strictly increasing since \(\displaystyle \log\left(\frac{a_{n+1}}{a_{n}}\right)=a_{n}\cdot \log(\alpha)>0\Rightarrow a_{n+1}>a_{n}\) and by the D'Alembert convergence criterion we get
\[\overline{\lim_{n\to\infty}}\left|\frac{\frac{a_{n+1}}{(n+1)^{\beta}}}{\frac{a_{n}}{n^{\beta}}}\right|=\overline{\lim_{n\to\infty}}\left|\frac{a_{n+1}}{a_{n}}\left(1+\frac{1}{n}\right)^{\beta}\right|>1\]
and so the series is divergent for all \(\beta\).