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Suppose $X_n$ are independent Cauchy r.v.s. I'm trying to prove that $\limsup \log X_n/\log n = c$ almost surely for some constant $c$. I know that by Borel-Cantelli it suffices is prove that

$\sum\mathbb{P}(X_n\ge n^c)=\infty$ and $\sum\mathbb{P}(X_n\geq n^{c+\epsilon})<\infty$ $\forall\epsilon>0$

However $\mathbb{P}(X_n\geq n^c)=\frac{1}{2}-\frac{1}{\pi}\arctan(n^c)$ and I know no way of evaluating this sum. Are there any convergence tests I should use that I've managed to miss? Or have I made a silly error somewhere? Thanks!

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For large $x$ $$\arctan(x) = \frac{\pi}{2} - \frac{1}{x} + \frac{1}{3x^3} - \frac{1}{5x^5} + O\left(\frac{1}{x^7}\right)$$ –  Henry May 28 '12 at 23:38

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up vote 3 down vote accepted

$$ \frac12-\frac1\pi\arctan(n^c)=\frac1\pi\arctan\left(\frac1{n^c}\right)\sim\frac1\pi\frac1{n^c} $$

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