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I'm working on a problem dealing with convergence in distribution of a sequence of random variables (in particular, $1/n$ times the maximum of $n$ Cauchy random variables) and have ended up with needing to take the limit of $$\left(\frac{\text{tan}^{-1}(nt)}{\pi}+\frac{1}{2} \right)^n $$ as $n\rightarrow \infty$.

I tried rewriting it, for example, as $$\frac{1}{2^n}\left(\frac{n\cdot2\text{tan}^{-1}(nt)/\pi}{n}+1 \right)^n $$ hoping to get something that looks like $e$, but haven't really been able to come up with anything.

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2 Answers 2

up vote 1 down vote accepted

The Taylor series expansion at infinity for $\tan^{-1}(nt)$ looks to be

$\frac{\pi}{2} - \frac{1}{nt} + O(\frac{1}{n^3})$.

for $t \geq 1.$ So plugging in to your equation should give you $(1 - \frac{1}{n\pi t})^n$, and taking the limit gives you $e^{-\frac{1}{\pi t}}$.

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Good call. Thanks! –  pedrosuavo Jan 16 '13 at 4:22

If you take the log of this function, this limit diverges for every t, which implies this limit diverges too.

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Hmmm. Wolfram Alpha seems to say it converges, at least for t = 1: wolframalpha.com/input/… –  Bitrex Jan 16 '13 at 3:42

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