I'm currently reading a book Analysis I by Amann and Escher. They define a cluster point of a sequence in a metric space as follows:
Let $(X,d)$ be a metric space. $a \in X$ is a cluster point of a sequence $(x_n)$ in $X$ if every neighborhood of $a$ contains infinitely many terms of the sequence.
It can be proven the $a$ being a cluster point is equivalent to the following two conditions:
For each neighborhood $U$ of $a$ and $m \in \mathbb{N}$ there is $n \geq m$ such that $x_n \in U$,
for each $\epsilon > 0$ and $m \in \mathbb{N}$, there is $n \geq m$ such that $x_n \in B(a,\epsilon)$.
It can be proved that for a sequence $(x_n)$ in $\mathbb{R}$ which is bounded above, $\overline{\lim} x_n = \lim_{n \in \mathbb{N}}\sup_{k \geq n} x_k$ is the greatest cluster point of the $(x_n)$.
What I don't understand is what property of a greatest cluster point is used in this proof (I only present those parts of the theorem and the proof which a relevant here):
Proposition. Let $\sum x_k$ be a series in a Banach space $(E,||\cdot||)$ and let $\alpha = \overline{\lim} \sqrt[k]{||x_k||}$. If $\alpha < 1$, then $\sum x_k$ converges absolutely.
Proof. If $\alpha < 1$, then the interval $(\alpha,1)$ is not empty and we can choose some $q \in (\alpha,1)$. We know that $\alpha$ is the greatest cluster point of the sequence $(\sqrt[k]{||x_k||})$. Hence there is some $K$ such that $\sqrt[k]{||x_k||} < q$ for all $k \geq K$...
I can't see why there would be such $K$. Indeed, for each $m \in \mathbb{N}$ there is $n \geq m$ such that $\sqrt[k]{||x_k||} < \alpha + (q - \alpha) = q$. It's clear that one should somehow use the property of $\alpha$ being the greatest cluster point, but I don't see how.