Limit involving arctan

Question

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.

Answer 1

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 $e^{-\frac{1}{\pi t}}$.

Answer 2

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