# If $X_j \sim [0, \sigma_j^2]$ is uniformly bounded, how show $\dfrac{\max_{1 \leq j\leq n} \sigma_j}{\sqrt{Var\left(\sum_{j=1}^n X_j\right)}} \to 0$?

Suppose that $|X_j| \leq C <\infty$ is a uniformly bounded sequence for $j=1,2,\ldots$ with $X_j \sim [0, \sigma_j^2]$ and that $X_j$ are independent. I would like to show the UAN condition for central limit theorems, in that:

$$\dfrac{\max_{1 \leq j\leq n} \sigma_j}{\sqrt{Var\left(\sum_{j=1}^n X_j\right)}} \to 0$$

I am not sure how to bound this. The best I came up with still involves $n$ in the numerator. Does anyone have any ideas? Thanks.

Using independence and the fact that all the random variables are centered, to question reduces to the following: does the convergence $$\tag{*}\lim_{n\to +\infty} \frac{\max_{1\leqslant j\leqslant n}\sigma_j^2}{\sum_{i=1}^n\sigma_i^2 } =0$$ hold? If the series $\sum_{i=1}^{ +\infty} \sigma_i^2$ diverges, then we are done, since $\sigma_j^2\leqslant C^2$ for each $j$. Otherwise, (*) does not hold if the series $\sum_{i=1}^{ +\infty} \sigma_i^2$ converges because there exists $N$ such that $\sigma_N\neq 0$ (otherwise we would divide by zero) and for $n\geqslant N$, $$\frac{\max_{1\leqslant j\leqslant n}\sigma_j^2}{\sum_{i=1}^n\sigma_i^2 } \geqslant \frac{ \sigma_N^2}{\sum_{i=1}^n\sigma_i^2 }$$ an the right hand side converges to a positive constant.