I am reading about convergence rates in the central limit theorem and have a question regarding Esseen's inequality.
Suppose $(X_j)_{j\in\mathbb{N}}$ is an independent sequence of random variables such that $\mathbb{E}X_j=0$ and $\mathbb{E}\lvert X_j\rvert^3<\infty$ for $j\in\mathbb{N}$. Define $\sigma^2_j:=\mathbb{E}X_j^2$ for $j\in\mathbb{N}$. As a reminder:
Esseen's inequality:. Let $n \in\mathbb{N}$. Put $B_n:=\sum_{j=1}^n\sigma_j^2$ and $F_n(x):=\mathbb{P}\left(B_n^{-1/2}\sum_{j=1}^n X_j<x\right)$ for $x\in\mathbb{R}$. Then for some absolute constant $c$ $$\sup_{x\in\mathbb{R}}\left\lvert F_n(x)-\Phi(x)\right\rvert \leq cB_n^{-3/2}\sum_{j=1}^n \mathbb{E}\lvert X_j\rvert^3 \tag{1},$$ where $\Phi$ denotes the cdf of the standard normal distribution.
Is then also the following true (with or without additional assumptions)?
Esseen's inequality for random series: Assume additionally that all the (random) series in the following converge and put $F(x):=\mathbb{P}\left(\left(\sum_{j=1}^\infty\sigma_j^2\right)^{-1/2}\sum_{j=1}^\infty X_j<x\right)$ for $x\in\mathbb{R}$. Then, $$\sup_{x\in\mathbb{R}}\left\lvert F(x)-\Phi(x)\right\rvert \leq c\left(\sum_{j=1}^\infty\sigma_j^2\right)^{-3/2}\sum_{j=1}^\infty \mathbb{E}\lvert X_j\rvert^3,$$ where $c$ is the same constant as above.
Do we need that $F_n$ converges uniformly to $\Phi$, in order to interchange the limit and supremem on the LHS, when taking the limit as $n\to\infty$ in (1)?
EDIT: To clarify, I am additionally assuming here that $\sum_{j=1}^\infty\sigma_j^2<\infty$, $\sum_{j=1}^\infty \mathbb{E}\lvert X_j\rvert^3<\infty$, and the random series $\sum_{j=1}^\infty X_j$ converges almost surely.
Fix a positive $\varepsilon$ and let $A_n$ be the event $$ A_n:=\left\{\left\lvert \left(\sum_{j=1}^\infty\sigma_j^2\right)^{-1/2}\sum_{j=1}^\infty X_j-B_n^{-1/2}\sum_{j=1}^n X_j \right\rvert \gt \varepsilon\right\}.$$ Then for all $x\in\mathbb R$, $$ F(x)-\Phi(x)\leqslant \mathbb{P}\left(\left(\left(\sum_{j=1}^\infty\sigma_j^2\right)^{-1/2}\sum_{j=1}^\infty X_j<x\right)\cap A_n^c\right)+\mathbb P(A_n)-\Phi(x) $$ and the inclusion $$ \left\{ \left(\sum_{j=1}^\infty\sigma_j^2\right)^{-1/2}\sum_{j=1}^\infty X_j<x\right\}\cap A_n^c \subset \left\{B_n^{-1/2}\sum_{j=1}^n X_j<x+\varepsilon \right\} $$ hold hence $$ F(x)-\Phi(x)\leqslant F_n(x+\varepsilon)-\Phi(x+\varepsilon)+ \Phi(x+\varepsilon)-\Phi(x)+\mathbb P(A_n).$$ Thus, Esseen's inequality gives that for all $n$, $$ F(x)-\Phi(x)\leqslant cB_n^{-3/2}\sum_{j=1}^n \mathbb{E}\lvert X_j\rvert^3+ \sup_{x\in\mathbb R} \left(\Phi(x+\varepsilon)-\Phi(x)\right)+\mathbb P(A_n) $$ and letting $n$ going to infinity gives $$ F(x)-\Phi(x)\leqslant c\left(\sum_{j=1}^\infty\sigma_j^2\right)^{-3/2}\sum_{j=1}^\infty \mathbb{E}\lvert X_j\rvert^3+ \sup_{x\in\mathbb R} \left(\Phi(x+\varepsilon)-\Phi(x)\right) . $$ The later supremum is reached for $x=-\varepsilon/2$ and is $\mathbb P\left(-\varepsilon/2\leqslant N\leqslant \varepsilon/2\right)$, where $N$ has a standard normal distribution and this goes to $0$ as $\varepsilon$ goes to $0$.
It remains to find an upper bound for $\Phi(x)-F(x)$. Here again, intersect with $A_n$ and control the difference between $F(x)$ and $F_n(x-\varepsilon)$.