As in Nate's answer, we are interested in iterating the function $$G(y) := \frac{ \int_{y}^{\infty} x e^{- x^2} dx}{\int_{y}^{\infty} e^{- x^2} }.$$
The numerator is $e^{-y^2}/2$ (elementary). The denominator is $e^{-y^2}/2 \cdot y^{-1} \left( 1-(1/2) y^{-2} + O(y^{-4}) \right)$ (see Wikipedia). So $G(y) = y + (1/2) y^{-1} + O(y^{-3})$.
Set $z_n = \mu_n^2$. Then $$z_{n+1} = (\mu_n+\mu_n^{-1}/2 + O(\mu_n^{-3}))^2 = \mu_n^2 + 1 + O(\mu_{n}^{-1}O(\mu_{n}^{-2}) = z_n + 1 + O(z_n^{-1/2}).$$O(z_n^{-1}).$$ So z_n \approx n and we see that \mu_n \to \infty like \sqrt{n}. I haven't checked the details, but I think you should be able to get something like \mu_n = n^{1/2} + O(1). 1 As in Nate's answer, we are interested in iterating the function $$G(y) := \frac{ \int_{y}^{\infty} x e^{- x^2} dx}{\int_{y}^{\infty} e^{- x^2} }.$$ The numerator is e^{-y^2}/2 (elementary). The denominator is e^{-y^2}/2 \cdot y^{-1} \left( 1-(1/2) y^{-2} + O(y^{-4}) \right) (see Wikipedia). So G(y) = y + (1/2) y^{-1} + O(y^{-3}). Set z_n = \mu_n^2. Then $$z_{n+1} = (\mu_n+\mu_n^{-1}/2 + O(\mu_n^{-3}))^2 = \mu_n^2 + 1 + O(\mu_{n}^{-1}) = z_n + 1 + O(z_n^{-1/2}). So $z_n \approx n$ and we see that $\mu_n \to \infty$ like $\sqrt{n}$.
I haven't checked the details, but I think you should be able to get something like $\mu_n = n^{1/2} + O(1)$.