To help avoid any misunderstanding that may arise for readers of this question, let me say that when understood in the usual sense, there are no nontrivial convergent sequences or series at all in the hyperreal numbers (and neither in the surreal numbers).
The reason is that the hyperreal numbers are usually understood to be countably saturated. This has a model-theoretic definition, meaning that every finitely realized type with countably many parameters is realized. But in the case of ordered fields, it can be expressed as the following property: whenever $$x_0\leq x_1\leq x_2\leq\cdots y_2\leq y_1\leq y_0$$ with $x_n<y_n$ for all $n\in\mathbb{N}$, then there is a number $z$ in between $$x_0\leq x_1\leq x_2\leq\cdots z \cdots y_2\leq y_1\leq y_0.$$
This saturation property prevents any nontrivial convergent sequence $x_n\to x$, unless it is eventually constant, since there will be various numbers $z$ in between, on each side.
Indeed, every countable set of hyperreal numbers is discrete in the hyperreal order. For this reason, one cannot apply any of the usual treatment of sequences and series from the real numbers in the hyperreal field.
Meanwhile, there is nevertheless a robust theory of sequences and series in the hyperreals, but using sequences and series indexed instead by $\mathbb{N}^*$, the nonstandard natural numbers, instead of merely $\mathbb{N}$. With this change, one can form hyperreal analogues of all the familiar sequences and series in the reals. The reason is that by the transfer principle, every assertion made in the real numbers about any such sequence or series is also true exactly the same in the nonstandard realm.
Thus, the radius of convergent of the nonstandard analogue of a power series remains the same as what it was.
For example, the series $$\sum_{n\in\mathbb{N}}\frac 1{2^n}=\frac12+\frac14+\cdots+\frac1{2^n}+\cdots$$ does not converge in the hyperreals, since one is using the natural numbers only. But the much longer (uncountably so) series $$\sum_{n\in\mathbb{N}^*}\frac 1{2^n}=\frac12+\frac14+\cdots+\frac1{2^n}+\cdots+\cdots=1$$ does converge to $1$.