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For your first question, the answer is yes, and I don't understand why it isn't better known since all the classical proofs of the central limit theorem generalize easily to that setting. See this section of the Wikipedia page on the central limit theorem.

Added: I overstated slightly, since one needs a slightly stronger condition than just $\sigma(S_n) \to \infty$, and the classical proofs don't easily generalize to give the most general conditions. But in the OP's setting, as coudy points out, Lindeberg's condition implies that it's enough to have $\sigma(S_n)/\sqrt{n} \sigma(S_n) \to \infty$ (whereas if the $X_i$ were identically distributed we would of course have $\sigma(S_n) = n\sigma(X)$).\sqrt{n}\sigma(X)$). For your second question, even under uniform boundedness,$c_n$only goes to zero like$n^{-1/2}$in general. See for example this paper. 4 Rollback to Revision 2 For your first question, the answer is yes, and I don't understand why it isn't better known since all the classical proofs of the central limit theorem generalize easily to that setting. See this section of the Wikipedia page on the central limit theorem. Added: I overstated slightly, since one needs a slightly stronger condition than just$\sigma(S_n) \to \infty$, and the classical proofs don't easily generalize to give the most general conditions. But in the OP's setting, Lyapunov's Lindeberg's condition implies that it's enough to have$\sigma(S_n)/n^{(1/2)-\varepsilon} \sigma(S_n)/\sqrt{n} \to \infty$(whereas if the$X_i$were identically distributed we would of course have$\sigma(S_n) = \sqrt{n}\sigma(X)$).n\sigma(X)$).

For your second question, even under uniform boundedness, $c_n$ only goes to zero like $n^{-1/2}$ in general. See for example this paper.

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For your first question, the answer is yes, and I don't understand why it isn't better known since all the classical proofs of the central limit theorem generalize easily to that setting. See this section of the Wikipedia page on the central limit theorem.

Added: I overstated slightly, since one needs a slightly stronger condition than just $\sigma(S_n) \to \infty$, and the classical proofs don't easily generalize to give the most general conditions. But in the OP's setting, Lindeberg's Lyapunov's condition implies that it's enough to have $\sigma(S_n)/\sqrt{n} \sigma(S_n)/n^{(1/2)-\varepsilon} \to \infty$ (whereas if the $X_i$ were identically distributed we would of course have $\sigma(S_n) = n\sigma(X)$).\sqrt{n}\sigma(X)$). For your second question, even under uniform boundedness,$c_n$only goes to zero like$n^{-1/2}\$ in general. See for example this paper.

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