When proving convergence of power series, one encounters the following term:
$$\left( 1 + \delta \right)\frac{\sqrt[k]{|a_k|}}{\lim\sup_{n\to\infty}\sqrt[n]{|a_n|}}$$
where $\delta>0$, and the claim is that this term is greater than one from the definition of $\lim\sup_{n\to\infty} a_n=\lim_{n\to\infty} \sup_{k\geq n}(a_k)$. But I do not see that.
Note: The statement and proof of divergence goes as this:
The series $\sum_{k=0}^\infty a_k z^k$ diverges on $[z\in\mathbb{C}: |z|>R]$ with $R=\frac{1}{\lim\sup_{n\to\infty}\sqrt[n]{|a_n|}}$.
Let now $R\in(0,\infty)$, then we can write $|z|=(1+\delta)R$, with $\delta>0$, one has:
$$ |a_k z^k| = |a_k| |z|^k = \left( \left(1+\delta\right) \frac{\sqrt[k]{|a_k|}}{\lim\sup_{n\to\infty}\sqrt[n]{|a_n|}}\right)^k $$
And now one is supposed to use the fact that the inside of the bracket is greater than 1 for infinitely many indices $k$.
Note, that there is no assumption on the $[a_k]$ other than $[a_k]\in\mathbb{C}$.
PS: Can someone put in comments, how to write curly braces here? Standard escaping with backslash is not working.
Let $L:=\limsup_{n\to\infty}\sqrt[n]{|a_n|}$ and $L':=\frac L{1+\delta}.$ Since $L'<L,$ by definition of the $\limsup,$ $\sqrt[k]{|a_k|}>L'$ for infinitely many indices $k.$ For these $k$'s, we have $$(1+\delta)\frac{\sqrt[k]{|a_k|}}{\lim\sup_{n\to\infty}\sqrt[n]{|a_n|}}=\frac{\sqrt[k]{|a_k|}}{L'}>1.$$