The binomial series of $(1-x^n)^{\frac{1}{n}}$, where $n$ is a positive integer, converges absolutely to $(1-x^n)^{\frac{1}{n}}$ for $x\in[0,1]$. The binomial series expansion is $$(1-x^n)^{\frac{1}{n}}=\sum_{k=0}^{\infty}{\binom{\frac{1}{n}}{k}(-x^n)^k}.$$ So suppose I define $$f_N(x)=\sum_{k=0}^{N}{\binom{\frac{1}{n}}{k}(-x^n)^k}.$$ By absolute convergence, $\forall x\in[0,1]$, we have $$N>M(x,\epsilon)\Rightarrow\left|(1-x^n)^{\frac{1}{n}}-f_N(x)\right|<\epsilon,$$ where $M(x,\epsilon)$ is the smallest (integer) $M$ that is valid, and is a function of $x$ and $\epsilon$. For any fixed $\epsilon$, $M(x)$ is a function $[0,1]\to\mathbb{Z}$. I want to show that for any $\epsilon$, $M(x)$ attains a supremum over $[0,1]$. How would I go about doing this?
If $M(x)$ was continuous, this would be simple since any continuous function on a compact set is bounded. But $M(x)$ is not necessarily continuous, and likely has jump discontinuities since it outputs positive integers. Perhaps a step would be to show local continuity? That if $x\approx y$, then $M(x)\approx M(y)$. Though I'm at a dead end here as well, and cannot properly phrase my $\epsilon$-$\delta$ argument.
For completeness, I will provide a solution that was given on a Facebook group.
Fix an $\epsilon>0$.
Since $(1-x^n)^{\frac{1}{n}}$ and $f_N(x)$ are continuous, we have that their difference is continuous, so for any $x\in[0,1]$, we can find some neighborhood of $x$, an open set $U$ containing $x$, such that $\forall y\in U$, we have $\left|(1-y^n)^{\frac{1}{n}}-f_N(y)\right|<\epsilon$.
Doing this for every $x\in[0,1]$ gives us a covering of $[0,1]$. But since $[0,1]$ is compact, by the Heine-Borel theorem, from that covering we can extract a finite subcovering of $[0,1]$. The open sets that remain in our subcovering correspond to a finite set $S=\{x_i\}$ with $x_i\in[0,1]$. Hence $M(x)\leq\max_{x_i\in S}{M(x_i)}$, and we are done. $\square$
Remark: This is equivalent to showing that the $f_N$ converge uniformly over $[0,1]$. So the problem can be solved by showing this instead.