# Root test and $\lim\sup$

I see how to define $$\lim \sup x_n$$ for a sequence $$(x_n)$$ in $$\mathbb{R}$$ which is bounded above: as a limit of a decreasing sequence $$(\sup_{k \geq n} x_k)$$ of real numbers. Since the extended real line $$\overline{\mathbb{R}}$$ is not even a metric space, the only way I see to define $$\lim\sup x_n$$ for sequences $$(x_n)$$ which are not bounded above is as $$\sup E$$ where $$E$$ is the set of all cluster points of $$(x_n)$$ or, equivalently, the set of limits of all convergent subsequences of $$(x_n)$$, but this seems to be useless in way that it is hard to compute: even if we prove that if concides with $$\lim_{n\to\infty} \sup_{k\geq n} x_k$$ in case when $$(x_n)$$ is bounded above, we wouldn't have the computational machinery of the theory of limits in case where it is not bounded, so it wouldn't be useful in applications such as the root test.

The root test is supposed to claim that

For a series $$\sum x_k$$ in a Banach space $$(E,||\cdot||)$$, setting $$\alpha = \lim\sup\sqrt[k]{||x_k||}$$, $$\alpha < 1$$ implies that $$\sum x_k$$ converges absolutely, while $$\alpha > 1$$ implies that $$\sum x_k$$ diverges.

To define $$\lim\sup\sqrt[k]{||x_k||}$$ rigorously and to be able to use the computational machinery of the theory of limits, can we restrict the criterion to series $$\sum x_k$$ such that the sequence $$(||x_k||)$$ is bounded above? After all, it being not bounded means that it diverges, hence, in particular, it doesn't converge to $$0$$, hence $$\sum x_k$$ diverges a priori.

However, I have doubts that setting such constraints as $$(||x_k||)$$ being bounded above reduces the usefullness of a root test. Is there another way, or should it be fine?

The superior limit of a sequence $$(x_n)_{n\in\mathbb N}$$ which is not bounded above can simply be defined as $$\infty$$. It will still be the supremum (in $$\overline{\mathbb R}$$) of all cluster points.
And it is not hard to to turn $$\overline{\mathbb R}$$ into a metric space. You consider the bijection$$\begin{array}{rccc}f\colon&\overline{\mathbb R}&\longrightarrow&[-1,1]\\&x&\mapsto&\begin{cases}\frac x{1+\lvert x\rvert}&\text{ if }x\in(-1,1)\\\pm1&\text{ if }x=\pm\infty\end{cases}\end{array}$$and you define the distance $$d$$ in $$\overline{\mathbb R}$$ by $$d(x,y)=\bigl\lvert f(x)-f(y)\bigr\rvert$$.
Finally, yes, if $$\bigl(\lVert x_n\rVert\bigr)_{n\in\mathbb N}$$ is unbounded, then you can just say that the series $$\sum_{n=1}^\infty x_n$$ diverges and that's it.