Suppose $Y:=(Y_j)_{j\in \mathbb N}$ is a sequence of indepedent random variables where $Y_j$ has probability mass function $f_j(y)=(2j+1)^{-1}$ for $y\in\{k/j\}_{k=-j}^j$. Show that $$S_n:=\sum_{j=1}^n \frac {Y_j}{\sqrt{n}}$$ converges in distribution as $n\to\infty$ and determine the limit.
My attempt:
$$\mathbb E(Y_j)=0$$ by symmetry.
$$Var(Y_j)=\sum_{k=-j}^j \frac{k^2}{j^2(2j+1)}=\frac{j+1}{3j}$$ then $$\sum_{j=1}^nVar(Y_j)=\sum_{j=1}^n\frac{j+1}{3j}=\frac n3+\frac 13 \sum_{j=1}^n\frac 1j$$ how do I proceed from the above, especially the extra sum, to claim the following that $$S_n\sim N\left(0,\frac 13\right)$$ which version of CLT is this?
I believe the $\frac{1}{\sqrt{n}}$ factor in the definition of $S_n$ should make the variance come out as needed when taking the limit of n going to infinity. It looks like it cancels out the first $n$ leaving $\frac{1}{3}$, and the extra sum will be multiplied by 0. Have you taken that into account? I hope this helps.