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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?

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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}

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