Given $r$ computed as the "mean resultant vector length" from a sample from the von Mises–Fisher distribution in $n$ dimensions, the $\kappa$ solving $$r=A_n(\kappa)=\frac{I_{n/2}(\kappa)}{I_{n/2-1}(\kappa)}$$ is the maximum likelihood estimate of the concentration parameter $\kappa$. $r$ is in $[0,1)$, the range of $A_n(\kappa)$ for $\kappa\in[0,\infty)$. $I_n(x)$ is the modified first-kind Bessel function.
I wrote an Observable notebook on the von Mises–Fisher, including a section on parameter estimation. Sra (2011) gives several ways of computing $A_n^{-1}$ and I used Newton's method in the notebook, but before that I tried using series expansions, which led me to the question below.
Despite $\lim_{r\to1^-}A_n^{-1}(r)=+\infty$, numerical calculations show that the absolute error of Banerjee et al.'s approximation (given in the Sra paper) $$A_n^{-1}(r)\approx\frac{r(n-r^2)}{1-r^2}=B_n(r)$$ is uniformly bounded over $[0,1)$ for all $n\ge2$, with $$\lim_{r\to1^-}(A_n^{-1}(r)-B_n(r))=-\frac12$$ independent of $n$.
How can this last limit be shown rigorously?
It would be nice if an exact series expansion of $A_n^{-1}(r)$ about $r=1$ can be obtained to prove the limit immediately. By series reversion it is easy to get expansions about $r=0$, e.g. $$A_2^{-1}(r)=2x+x^3+\frac{5x^5}6+\frac{19x^7}{24}+\frac{143x^9}{180}+\cdots$$
When $r$ is close to $1$, $\kappa$ is large and positive. Hence, one has to use the large-$\kappa$ asymptotics of the modified Bessel function (cf. $(10.40.1)$). In this way, we get $$ r = 1 - \frac{{n - 1}}{{2\kappa }} + \frac{{(n - 1)(n - 3)}}{{8\kappa ^2 }} + \frac{{(n - 1)(n - 3)}}{{8\kappa ^3 }} + \mathcal{O}\!\left( {\frac{1}{{\kappa ^4 }}} \right). $$ Inversion and algebraic manipulation yields \begin{align*} \kappa & = \frac{{n - 1}}{{2(1 - r)}} + \frac{{3 - n}}{4} + \mathcal{O}((1 - r)) \\ & = \frac{{r(n - r^2 )}}{{1 - r^2 }} - \frac{{2(r - 1)^2 + 6(r - 1) + (5 - n)}}{{2(r + 1)}} + \frac{{3 - n}}{4} + \mathcal{O}((1 - r)) \\ & = \frac{{r(n - r^2 )}}{{1 - r^2 }} - \frac{1}{2} + \mathcal{O}((1 - r)), \end{align*} which is the desired result.