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Suppose $X_1, X_2, \dots$ are independent random variables, with $X_k \sim N(0,2^{k-1})$ for each $k \in \mathbb{N}$, i.e., each $X_k$ is normally distributed with mean $0$ and variance $2^{k-1}$. For $n \in \mathbb{N}$, define $$S_n = \sum_{k=1}^n X_k, \quad s_n^2 = \sum_{k=1}^n \operatorname{Var}(X_k).$$ How do I show that $\frac{S_n}{s_n}$ converges in distribution to $N(0,1)$ as $n \to \infty$?

Initially, I tried to prove this by Lindeberg's condition: for any $\varepsilon > 0$, $$\frac{1}{s_n^2} \sum_{k=1}^n E[X_k^2 \mathbf{1}_{\{|X_k| > \varepsilon s_n\}}] \to 0.$$ However, as @kimchi lover pointed out, this condition is does not actually hold for this problem: an implication of Lindeberg's condition is that $\max_{k=1,\dots,n} \frac{\operatorname{Var}(X_k)}{s_n^2} \to 0$, but $$\max_{k=1,\dots,n} \frac{\operatorname{Var}(X_k)}{s_n^2} = \frac{2^{n-1}}{2^n - 1} \to \frac{1}{2} \neq 0.$$ This implies that the Lindeberg condition cannot hold for $(X_k)_{k=1}^\infty$.

Are there other methods to show that $\frac{S_n}{s_n}$ converges in distribution to $N(0,1)$? Or is this proposition not true in the first place?

Edit: I thought of a simple way to solve this using basic notions about normal distribution, but I'm not sure if this is the right way. This method only relies on the fact that the sum and scalar multiples of independent normally distributed random variables also has the normal distribution. Hence $\frac{S_n}{s_n}$ is normally distributed with mean $0$ and variance $$\frac{1}{s_n^2} \operatorname{Var}(S_n) = \frac{1}{s_n^2} \sum_{k=1}^n \operatorname{Var}(X_k) = 1.$$ Thus $\frac{S_n}{s_n} \sim N(0,1)$ for each $n$, so obviously $\frac{S_n}{s_n} \to N(0,1)$ in distribution. Is this line of reasoning correct? It seems to be almost too simple, so I might have misunderstood something.

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The attempt you give in the edit is correct. Probably the goal was to give an example where the central limit theorem holds but not the Lindeberg's condition.

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