# Limit supremum and differentiability of power series

There is this part in Newman and Bak's Complex Analysis where he justifies via lim sup, that the series obtained by differentiating a convergent series has the same radius of convergence as the former one. It goes:

Given a sequence $$\{C_n\}$$ in $$\mathbb{C}$$ such that $$\limsup\limits_{n\rightarrow\infty} |C_n|^{1/n}$$ exists, we have

$$\limsup\limits_{n\rightarrow\infty}|nC_n|^{1/(n-1)} = \limsup\limits_{n\rightarrow\infty}\left(|nC_n|^{1/n}\right)^{n/(n-1)} = \limsup\limits_{n\rightarrow\infty} |C_n|^{1/n}.$$

I just cannot understand the second equality. Before rewriting the exponent, we can use the fact that $$\lim\limits_{n\to\infty}|n|^{1/(n-1)}=1$$ and write

$$\limsup\limits_{n\rightarrow\infty}|nC_n|^{1/(n-1)} = \lim\limits_{n\to\infty}|n|^{1/(n-1)}\limsup\limits_{n\rightarrow\infty} |C_n|^{1/(n-1)} = \limsup\limits_{n\rightarrow\infty} |C_n|^{1/(n-1)} = \limsup\limits_{n\rightarrow\infty} |C_n|^{1/n}$$

How does rewriting the exponent that way help us?

I think that what the authors have in mind is this: for any sequence $$(a_n)_{n\in\Bbb N}$$ of numbers greater than $$0$$, we have$$\limsup_na_n=\limsup_na_n^{n/(n-1)},$$and therefore, since$$\limsup_n\left|nC_n\right|^{1/n}=\limsup_n\left|C_n\right|^{1/n},$$you have\begin{align}\limsup_n\left(\left|nC_n\right|^{1/n}\right)^{n/(n-1)}&=\limsup_n\left|nC_n\right|^{1/n}\\&=\limsup_n\left|C_n\right|^{1/n}.\end{align}