I am looking for the limit, as $n\rightarrow\infty$, for the formula $$ \mathrm{erf}^{-1}(2^{-1/n})-\frac{1}{c}\ln(-\log_2\beta) [\mathrm{erf}^{-1}(4^{-1/n})-\mathrm{erf}^{-1}((4/3)^{-1/n})] \enspace, $$ in which $c=-\ln\log_4 4/3$ is a constant. $\beta$ is a probability and $n\rightarrow\infty$.
I can (informally) get $O(\sqrt{\ln(1/\beta)\ln n})$ using the fact that the first term in the series expansion at $n\rightarrow\infty$ of $\mathrm{erf}^{-1}((4/x)^{-1/n})$ is $\sqrt{Z(x)-\ln(\pi Z(x)-\pi\ln(\pi))}/\sqrt{2}$ in which $Z(x)=\ln(2n^2/\ln ^2(4/x))$. However, I am looking for an exact limit or at least big $\Theta$ of it.
Background (for the interested)
This is a sort of a bound on the maximum deviation (measured by the infinity norm) a multinomial random variable can have from its mean with probability at most $\beta$.
I use the marginal random variables of the components of the multinomial vector, which are binomial random variables. Then, I use the normal approximation to the binomial distribution. The absolute value of the difference between each such random variable and its mean thus follows the half-normal distribution (this is where the erf function comes from). I then approximate the maximum of those i.i.d. $n$ half-normal random variables (they are i.i.d. at some limit) by a Gumbel distribution $\mathrm{M}+\mathrm{IQR} \cdot G$ whose median (M) and interquantile range (IQR) match the maximum of the i.i.d. half-normals (whose CDF is simply the product of their CDFs). Here $G=G(c^{-1}\ln\ln 2,c^{-1})$ is the Gumbel distribution whose median and interquantile range are 0 and 1, respectively.
I hope I have understood your question right.
$\displaystyle \mathrm{erf}(x):=\frac{1}{\Gamma(1.5)}\int\limits_0^x e^{-t^2}dt$
$\displaystyle x=\mathrm{erf}(\mathrm{erf}^{-1}(x))=\frac{1}{\Gamma(1.5)}\int\limits_0^{\mathrm{erf}^{-1}(x)} e^{-t^2}dt$
$\displaystyle x^{-\frac{1}{n}}=\frac{1}{\Gamma(1.5)}\int\limits_0^{\mathrm{erf}^{-1}(x^{-\frac{1}{n}})} e^{-t^2}dt \enspace$ , $\enspace x>0$
$n\to\infty$ : $\displaystyle \int\limits_0^{\mathrm{erf}^{-1}(1)} e^{-t^2}dt=\Gamma(1.5)$ which means that $\mathrm{erf}^{-1}(1)$ is well defined
Therefore one has
$\displaystyle \lim_{n\to\infty}(\mathrm{erf}^{-1}(2^{-1/n})-\frac{1}{c}\ln(-\log_2\beta) [\mathrm{erf}^{-1}(4^{-1/n})-\mathrm{erf}^{-1}((4/3)^{-1/n})])= \mathrm{erf}^{-1}(1)$ .