(The following question is equivalent to the title)
Let $(x_n)\in \ell^1$ be a real sequence, and let $1<p<\infty$. There exists an $\alpha>0$ such that the sequence $(n^{\alpha}x_n^p)$ is still in $\ell^1$?
By trying with $x_n = n^{-1-\varepsilon}$, I found that necessarily $\alpha<p-1$, but then I couldn't use this information. Ideally, I would need an Holder inequality, but with $p<1$, and the only one I found is in the opposite direction. I also found that the "dual" (in a certain sense, since $\ell^p$ with $p<1$ is not normed) of $\ell^p$ is $\ell^\infty$, that would rule out all $\alpha >0$, but here I am not requiring continuity of the operator.
Let $x_{n}$ be such that $x_{2^k}=1/k^2$ and $x_j=0$ if $j$ is not of the form $2^k$ for some integer $k$. Then $\sum\lvert x_n\vert=\sum_k 1/k^2<+\infty$ but for all $p>1$ and $\alpha>0$, $$\sum_n n^\alpha \lvert x_n\rvert^p=\sum_k 2^{k\alpha}k^{-2p}=+\infty.$$