Multiplying $\ell^\alpha$ elements by $n^\varepsilon$

Let $1 \le \alpha < \beta < \infty$ and $(a_n)_n \in \ell^\alpha(\mathbb N)$. Does this imply that there exists an $\varepsilon > 0$ such that $(a_n n^\varepsilon) \in \ell^\beta(\mathbb N)$?

Let $\alpha=1$ and $\beta>1$. Denote $[\gamma]$ the greatest integer less than $\gamma\in\mathbb{R}$.

For each $n\in\mathbb{N}$ with $n=[e^{k}]$ for some $k\in\mathbb{N}$, define $a_{n}=1/k^{2}$. For those $n\ne[e^{k}]$ for any $k$, we define $a_{n}=0$. It is easy to see that $(a_{n})\in\ell^{1}$.

Assuming that for some $\varepsilon>0$, $(a_{n}n^{\varepsilon})\in\ell^{\beta}$, then the tail $a_{n}n^{\varepsilon}\rightarrow 0$ as $n\rightarrow\infty$. Therefore, $a_{n}n^{\varepsilon}<1$ eventually, then for $n=[e^{k}]$, it follows that \begin{align*} a_{n}n^{\varepsilon}\geq\dfrac{(e^{k}-1)^{\varepsilon}}{k^{2}}\rightarrow\infty \end{align*} as $k\rightarrow\infty$. This contradicts that $a_{n}n^{\varepsilon}<1$ eventually.