Let $a, a\_n \in \mathbb{C}$ ($n \ge 2$). Suppose we are given the Laurent series 

$$ \sum\_{n=2}^{\infty} a\_n(z - a)^{-n}.$$

I know that we can calculate the set on which this expression converges using Hadamard's formula. That is, we have that the series converges for $|z - a| > \frac{1}{R}$, where $\frac{1}{R} = \limsup \sqrt[n]{|a\_n|}$. In case $\limsup \sqrt[n]{|a\_n|} = 0$, the series converges for $|z-a|> 0$. And in case $\limsup \sqrt[n]{|a\_n|} = \infty$, the series is nowhere convergent.

To continue, suppose that we formally anti-differentiate our series to obtain

$$ \sum\_{n=1}^{\infty} \frac{-a\_{n+1}}{n}(z - a)^{-n}.$$

My question is, how does the radius of convergence of this new series relate to the old one? Maybe the new radius of convergence is potentially larger? I see that the question boils down to just being smart about Hadamard's formula. We need to compare $\limsup \sqrt[n]{|a\_n|}$ with $\limsup \sqrt[n]{\frac{|a\_{n+1}|}{n}}$, but this seems like a tricky bit of analysis. It's already clear to me that $\limsup \sqrt[n]{\frac{|a\_{n+1}|}{n}} \le \limsup \sqrt[n]{|a\_{n+1}|}$, so my hope is that I can somehow compare $\limsup \sqrt[n]{|a\_{n+1}|}$ with $\limsup \sqrt[n]{|a\_n|}$.

Hints or solutions are greatly appreciated.