I am working from Durrett's Probability: Theory and Examples, and I have encountered the following question:
Suppose that $X$ is normally distributed, and $b_n$ is defined by $P(X>b_n)=\frac1n$. Show that $b_n\sim (2\log n)^{1/2}.$
I'm confused about how this could be true. We know that $P(X>x)$ is asymptotic to $\frac{1}{\sqrt{2\pi}x}e^{-x^2/2}$, and so we should get
$$\lim_{n\to\infty}\frac{\frac1n}{\frac{1}{\sqrt{2\pi}b_n}e^{-b_n^2/2}}=1$$
Heuristically, this seems unlikely, right? Since if we just replace $b_n$ with $(2\log n)^{1/2}$ we get $$\lim_{n\to\infty}\frac{\frac1n}{\frac{1}{\sqrt{2\pi}(2\log n)^{1/2}}\cdot\frac1n}=1,$$
which is obviously wrong.
Even if we try to go through a more rigorous formulation, I'm just not seeing where a factor of $\sqrt{\pi}$ is going to come from.