Show that if $K\subset\ell^{1}$ is compact, then for all $\epsilon>0$, there exists $N$ such that $\sum_{j=N+1}^{\infty}|x_{j}|<\epsilon$ uniformly. This question is related to the online note I referred in this post: Describe all the compact subsets of $\ell^{1}$ -- General Result Proved, Example Needed.
The online note there missed a proof of one direction, which can be described as

If $K\subset \ell^{1}$ is compact, then we must have $$\lim_{N\rightarrow\infty}\sup_{\mathbf{x}\in K}\sum_{j=N+1}^{\infty}|x_{j}|=0.$$

This is saying that for all $\epsilon>0$, there exists $N$ (only depends on $\epsilon$) such that $$\sum_{j=N+1}^{\infty}|x_{j}| < \epsilon$$ for all $\mathbf{x}\in K$.
I tried to prove this but it turned out in the end that my $N$ depends on the choice of $\mathbf{x}$. My proof is as follows:
As it is compact, it is totally bounded and closed. Let $\mathbf{x}\in K$ and $\epsilon>0$, then by total boundedness, there exists $m\in\mathbb{N}$, $\mathbf{x}_{1},\cdots, \mathbf{x}_{m}\in X$ and $\gamma\in \{1,\cdots,m\}$ such that $$\sum_{i=1}^{\infty}|x_{i}-x_{\gamma, i}|<\epsilon.$$
Note that since $\mathbf{x}_{\gamma}\in X$, $\sum_{j=1}^{\infty}|x_{\gamma, j}|<\infty$, which implies that there exists $N$ large enough such that $\sum_{j=N+1}^{\infty}|x_{\gamma, j}|<\epsilon.$
Hence, we have $$\sum_{j=N+1}^{\infty}|x_{j}|=\sum_{j=N+1}^{\infty}|x_{j}-x_{\gamma, j}+x_{\gamma, j}|\leq \sum_{j=1}^{\infty}|x_{j}-x_{\gamma, j}|+\sum_{j=N+1}^{\infty}|x_{\gamma, j}|<\epsilon+\epsilon=2\epsilon.$$

The problem in my proof is that $N$ depends on the choice of $\mathbf{x}_{\gamma}$, and the choice of $\mathbf{x}_{\gamma}$ depends on the choice of my $\mathbf{x}$.
What should I do to fix this? or my proof is actually correct? Thank you!
 A: The definition of totally boundedness is different to what you used in your argument.
$K$ being totally bounded implies that for any $\epsilon > 0$, there exist $\mathbf x_1, \dots, \mathbf x_m \in K$ such that for all $\mathbf x \in K$, there exists a $\gamma \in \{1, \dots, m \}$ such that $\sum_{i=1}^{\infty}|x_{i}-x_{\gamma, i}|<\epsilon.$
The rest of your argument then works!
Concretely, fix an $\epsilon > 0$ and pick $\mathbf x_1, \dots, \mathbf x_m \in K$ such that for all $\mathbf x \in K$, there exists a $\gamma \in \{1, \dots, m \}$ such that $\sum_{i=1}^{\infty}|x_{i}-x_{\gamma, i}|<\epsilon$, and pick an $N$ large enough such that $\sum_{j=N+1}^{\infty}|x_{\gamma, j}|<\epsilon$ for all of these $\gamma$'s.
For any given $\mathbf x \in K$, let $\gamma$ be the particular $\gamma$ such that $\sum_{i=1}^{\infty}|x_{i}-x_{\gamma, i}|<\epsilon.$ Then
$$\sum_{j=N+1}^{\infty}|x_{j}|=\sum_{j=N+1}^{\infty}|x_{j}-x_{\gamma, j}+x_{\gamma, j}|\leq \sum_{j=1}^{\infty}|x_{j}-x_{\gamma, j}|+\sum_{j=N+1}^{\infty}|x_{\gamma, j}|<\epsilon+\epsilon=2\epsilon.$$
Thus the $N$ depends on the choice of $\epsilon$ (which it is allowed to), but the $N$ does not depend on the choice of $\mathbf x$.
