Let $(a_k)_{k \in \mathbb{N}}$ be some real sequence. Show that if $\limsup_{k \rightarrow \infty} \sqrt[k]{|a_k|} < 1$ then $\sum_{k=0}^\infty a_k$ converges absolutely.
I have some general questions about the limes supremum first.
$\lim_{k \rightarrow \infty} \sup (\sqrt[k]{|a_k|})$ is defined as $\lim_{k \rightarrow \infty} (\sup(\bigcup_{i=k}^\infty (\sqrt[i]{|a_i|}))$
How exactly is the union of the set to be interpreted? $\bigcup_{i=k}^\infty (\sqrt[i]{|a_i|})$ is the union of all $\sqrt[i]{|a_i|}$ for all i larger than some k, right?
Now the supremum of that union is the largest term of that set. So I think this supremum depends on the choice of k. Is the supremum of the union therfore loosely speaking a function of k? And if I let k go to infinity I get the limes supremum, correct?
Now in the proof I don't understand how that limit could be less than 1. Since $\forall c \gt 0 \in \mathbb{R}: \lim_{k=0}^\infty \sqrt[k]{c_k} = 1$ So how can $\lim_{k=0}^\infty \sqrt[k]{|a_k|} < 1$?
And I guess the proof somehow has to make use of the root test. But if I take $\sup(\bigcup_{i=k}^\infty (\sqrt[i]{|a_i|})$, which yields a term of the form $\sqrt[k_0]{|a_{k_0}|})$ and then I let $k_0 \rightarrow \infty$, and by assumption this limit is less than 1, does it immidiately follow by the root test that $\sum_{k=0}^\infty a_k$ converges absolutely?