Bounding the fractional parts of the $p^{\text{th}}$ roots of $n,n^2,...,n^{p-1}$ EDIT (August 9, 2021): I would like to ask a more general question. The original question that was fully answered is below the line.
For a positive real number $x$, denote the fractional part $x-[x]$ of $x$ by $\langle x \rangle$.
Let $\ell>0$ be an integer. Is
$$\Phi_{\ell} := \liminf_{n>0 \text{ not a } {\ell}^{\text{th}} \text{ power}} \ \sum_{k = 1}^{\ell^2} \left\langle n^{\frac{k}{\ell}}\right\rangle$$
equal to zero?

Fix a prime $p$. Is it known whether or not
$$\Theta_p := \liminf_{n>0 \text{ not a } p^{\text{th}} \text{ power}} \ \sum_{k = 1}^{p-1} \left\langle n^{\frac{k}{p}}\right\rangle$$
is zero?
 A: Let $n=m^p+1$ for some large enough $m$. For $0<k<p$ we then have $m^{kp}<n^k<m^{kp}+O(m^{(k-1)p})=m^{kp}(1+O(m^{-p}))$, so $$m^k<n^{k/p}<m^k(1+m^{-p})^{k/p}\leq m^k(1+O(m^{-p}))=m^k+o(1).$$ Hence $\langle n^{k/p}\rangle=o(1)$ as $m\to 0$ for each $0<k<p$, and in particular the $\liminf$ in your question is zero.
The $\liminf$ in the edited question also tends to zero, though to see this now we have to consider more terms of the Taylor expansion. Specifically, let us again pick $n=m^\ell+1$, where now $m$ to be divisible by a suitable number, to be specified later. We may assume $\ell\nmid k$ and pick $d$ such that $d\ell<k<(d+1)\ell$. Then we have
$$n^{k/\ell}=m^k(1+m^{-\ell})^{k/\ell}=m^k(1+a_1m^{-\ell}+a_2m^{-2\ell}+\dots+a_d m^{-d\ell}+O(m^{-(d+1)\ell}))\\
=m^k+a_1m^{k-\ell}+a_2m^{k-2\ell}+\dots+a_d m^{k-d\ell}+O(m^{k-(d+1)\ell}),$$
where $a_i$ are some rational numbers. If we pick $m$ divisible by all of their denominators, then the terms up to $a_d m^{k-d\ell}$ will be integers, and the $O(m^{k-(d+1)\ell})$ term will tend to zero since the exponent is negative. This at least shows that $n^{k/\ell}$ can be arbitrarily close to an integer. So see that the fractional part is small we also have to see that the error term is positive. This follows from the fact that the coefficient of the next term in the Taylor expansion, $a_{d+1}m^{k-(d+1)\ell}$, has coefficient
$$a_{d+1}=\binom{k/\ell}{d+1}=\frac{k/\ell(k/\ell-1)\dots(k/\ell-d)}{(d+1)!}$$
which is positive by our choice of $d$.
The same argument should work if you replace $\ell^2$ in the sum by any constant (possibly dependent on $\ell$, but not $n$).
