in $cs$, the space of all sequences $(a_n)$ such that the series ${\displaystyle \sum _{i=1}^{\infty }a_{n}}$ is convergent (possibly conditionally), the norm of a sequence is defined as ${\displaystyle \left\|(x_n)\right\|_{cs}=\sup _{n}\left|\sum _{i=1}^{n}x_{i}\right|}$. for $(a_n)\in cs$ we can also talk about an induced radius ${}_i{r}((a_n))$ defined as the radius of convergence for $\displaystyle \sum _{i=1}^{\infty }a_{n}x^n$ which is given by ${}_ir((a_n)) = \frac{1}{\limsup_{n\rightarrow\infty}{|a_n|^{\frac{1}{n}}}}$. My question then is whether the induced radius is a continuous function where it is a real number, that is if we know ${}_i{r}((a_n)) \in \mathbb{R}$
$$\forall \epsilon>0 \stackrel{?}{\exists} \delta>0 \ (\displaystyle \left\|(h_n)\right\|_{cs}<\delta \Rightarrow |{}_i{r}((a_n+h_n))-{}_i{r}((a_n))|<\epsilon)$$ sommewhat helpful source: https://en.wikipedia.org/wiki/Bs_space
This result does not seem to hold. As a counter example I choose for a given $\delta>0$
Note that $\|(h_n)\|\le\delta$ and using the comments we have $r((a_n)) = 3$. However, when we look at $r((a_n+h_n))$ we see that \begin{align} r((a_n+h_n)) &= \left(\limsup_{n\to\infty}|3^{-n}+(-1)^n\delta|^{\frac{1}{n}}\right)^{-1} \\ &= \left(\limsup_{n\to\infty}|\delta|^{\frac{1}{n}}\right)^{-1} \\ &= 1. \end{align} It follows that we have $r((a_n+h_n)) - r((a_n)) = 2$, regardless of the $\delta$ chosen.