Let $\{X_k\}_{k\geq 1}$ be a sequence of independent random variables with uniform distribution on $[-1,1]$. Compute $$\lim_{n\rightarrow \infty} P\left(\left|\sum_{k=1}^n \frac{1}{X_k}\right|>\frac{\pi n}{2}\right)$$
Actually, I don't have good idea to approach this one. Can anyone give me some idea to do this?
Let $Y_n$ be defined by
$$ Y_n = \frac{1}{n}\sum_{k=1}^{n} \frac{1}{X_k}. $$
Assuming that $Y_n$ converges in distribution to a continuous random variable $Y$, the limiting probability is $\Bbb{P}(|Y| > \pi/2)$, and this may be evaluated once we know the distribution of $Y$. We claim that this is indeed true:
Claim. $Y_n$ converges in distribution to a random variable $Y$ having the Cauchy distribution with the scale parameter $\gamma = \pi/2$: $$ f_Y(y) = \frac{1}{\pi\gamma} \cdot \frac{\gamma^2}{y^2 + \gamma^2}. $$
Assuming this claim, we have
$$ \lim_{n\to\infty} \Bbb{P}(|Y_n| > \gamma) = \Bbb{P}(|Y| > \gamma) = \int_{|y| > \gamma} f_Y(y) \, dy = \frac{1}{2}. $$
Proof of the Claim. We show that the characteristic function $\phi_{Y_n}(t)$ of $Y_n$ converges to $\phi_{Y}(t)$. Notice that $\phi_{Y_n}(t)$ is given by
$$ \phi_{Y_n}(t) = \phi_{1/X}(t/n)^n = \left( \int_{|x|<1} \frac{e^{it/nx}}{2} \, dx \right)^n = \left( 1 - \int_{|x|<1} \frac{1 - \cos(t/nx)}{2} \, dx \right)^n $$
The last integral can be simplified by the substitution $y = 1/nx$:
$$ \int_{|x|<1} \frac{1 - \cos(t/nx)}{2} \, dx = \frac{1}{n} \int_{1/n}^{\infty} \frac{1 - \cos(ty)}{y^2} \, dy. $$
Plugging this back and taking $n\to\infty$, we have
$$ \lim_{n\to\infty} \phi_{Y_n}(t) = \exp\left\{ - \int_{0}^{\infty} \frac{1 - \cos(ty)}{y^2} \, dy \right\}. $$
This integral is easily computed from the Dirichlet integral. Indeed, integration by parts yields
$$ \int_{0}^{\infty} \frac{1 - \cos(ty)}{y^2} \, dy = \int_{0}^{\infty} \frac{t\sin(ty)}{y} \, dy = \gamma|t|, $$
where $\gamma = \pi/2$ is the scale parameter of the distribution of $Y$. Therefore we have
$$ \lim_{n\to\infty} \phi_{Y_n}(t) = e^{-\gamma|t|} = \phi_Y(t). $$