Question: Let $(X_k)_{k \geq 1}$ be a sequence of independent random variables with uniform distribution on $[-1,1]$ Compute $\lim_{n \to \infty} \mathbb{P}(|\sum_{k=1}^{n} X_{k}^{-1}|> \pi n/2)$
I tried to solve it using characteristic function, but the calculation seems very complicated However, I got the characteristic function to be $1$ as $n$ approaches infinity, which doesn't really give any assumption about the distribution function.
Let $X$ be uniformly distributed in $[-1,1]$, then $$ \mathbb{P}(X^{-1}\leqslant x)= \begin{cases} -\frac1{2x}, & \hfill x\leqslant-1\hfill \\ \frac12, & -1\leqslant x\leqslant 1 \\ 1-\frac1{2x}, & \hfill x\geqslant1\hfill \end{cases} $$ and we easily find \begin{align} \varphi(t)&:=\mathbb{E}e^{itX^{-1}} =\int_1^\infty\frac{\cos tx}{x^2}\,dx \\&=\cos t-t\int_1^\infty\frac{\sin tx}{x}\,dx \\&=\cos t-\frac\pi2|t|+t\int_0^1\frac{\sin tx}{x}\,dx, \end{align} so that $\varphi(t)=1-\pi|t|/2+O(t^2)$ as $t\to 0$.
Now, if $Y_n=\frac1n\sum_{k=1}^n X_k^{-1}$, then $\mathbb{E}e^{itY_n}=\varphi^n(t/n)$, and inversion gives$$P_n:=\mathbb{P}\left(|Y_n|\leqslant\frac\pi2\right)=\frac1\pi\int_{-\infty}^\infty\frac{\sin(\pi t/2)}{t}\,\varphi^n\left(\frac tn\right)dt.$$
To take $n\to\infty$ here, we see that the above implies $\varphi^n(t/n)\to e^{-\pi|t|/2}$ (and this is a dominated convergence). Using $\int_{-\infty}^\infty=2\int_0^\infty$ and substituting $x=\pi t/2$, we then get $$ \lim_{n\to\infty}P_n=\frac2\pi\int_0^\infty\frac{\sin x}x\,e^{-x}\,dx=\frac12 $$ (the integral can be computed in a number of ways).