I came across this problem in our probability course:
Let $\{X_{n}\}$ be a sequence of independent random variables. For all n, $$P\{X_{n}=n\} = P\{X_{n}=-n\} = \frac{1}{2n^2}, \quad P\{X_{n}=1\} = P\{X_{n}=-1\} = \frac{1}{2} - \frac{1}{2n^2}$$ Denote $S_{n} = \sum_{j=1}^{n} X_{j}$. Show that $\frac{Var[S_{n}]}{n} \rightarrow 2$ as $n \rightarrow \infty$.
I tried to write down the characteristic function of $S_{n}$ explicitly, but it turned out to be an infinite product of some combination of cosines and stuff, so I couldn't use $E[S_{n}^2] = \phi^{''}_{S_{n}}(0)$ to compute the variance. But I don't see any other way of doing it.
You don't need characteristic functions; you can instead compute this directly. Notice that since $X_n$ is centred, $$Var(X_n) = E(X_n^2)= \sum x^2 P(X_n = x) = 2(n^2 \cdot \frac{1}{2n^2}) + 2 \left( \frac{1}{2} - \frac{1}{2n^2}\right) = 2 - \frac{1}{n^2}$$ By independence, $$Var(S_n) = \sum_{i=1}^n Var(X_i) = \sum_{i=1}^n \left( 2 - \frac{1}{i^2}\right)= 2n - \sum_{i=1}^n \frac{1}{i^2}$$ Recall that $\sum_{i=1}^\infty \frac{1}{i^2} = \frac{\pi^2}{6} < \infty$, so that $$\frac{Var(S_n)}{n} = 2 - \frac{1}{n} \sum_{i=1}^n \frac{1}{i^2} \to 2$$