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Well, as the title says, I'm trying to solve the following limit:

$$\lim_{n\to\infty}\left[\frac{1}{\pi}\arctan\left(\frac{nx}{\pi}\right)+\frac{1}{2}\right]^n.$$

This arised in the following context: let $(X_i)$ be a sequence of independent Cauchy random variables. Let $M_n=\max_{1\leq i\leq n}X_i$. I want to calculate the distribution to which $\pi M_n/n$ tends to. My approach went as follows:

$$F_{\pi M_n/n}(x) = \mathbb{P}(\pi M_n/n\leq x)=\mathbb{P}(M_n\leq nx/\pi)=(\mathbb{P}(X_1\leq nx/\pi))^n.$$ Which is equals to $$\left[\frac{1}{\pi}\arctan\left(\frac{nx}{\pi}\right)+\frac{1}{2}\right]^n.$$

Ѕᴀᴀᴅ
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Gabriel
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1 Answers1

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Note that for $t\not=0$, $$\arctan\left(t\right)+\arctan\left(\frac{1}{t}\right)=\text{sign}(t)\cdot \frac{\pi}{2}.$$ (see for example Proving that $\arctan(x)+\arctan(1/x)=\pm \pi/2$, could this line of reasoning possibly be correct?).

Hence for $x>0$, as $n\to+\infty$, $$\left[1-\frac{1}{\pi}\arctan\left(\frac{\pi}{nx}\right)\right]^n =\exp\left(n\ln\left(1-\frac{1}{\pi}\arctan\left(\frac{\pi}{nx}\right)\right)\right)\to e^{-1/x}.$$ Are you able/interested to evaluate the limit when $x\leq 0$?

Robert Z
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