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am trying to show the following:

ILet $\xi_1,\xi_2,...$ be a sequence of independent identically distributed random variables with zero mean and finite and nonzero variance. Prove that the distributions of$ \frac{\sum_{i=1}^{n} \xi_i}{(\sum_{i=1}^{n} \xi_i^2)^{\frac{1}{2}}}$converge weakly to $N(0, 1)$ distribution as $n \to \infty$.

I think this might be an application of the Lyapunov CLT formulation, but I'm not sure if it's straightforward how to show that the Lyapunov Condition holds. I'm not really sure how to proceed.

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    $\begingroup$ You can divide both the numerator and denominator by $\sqrt n\sigma$ (where $\sigma^2$ is the variance of $\xi_i$), use the common CLT for numerator and Slutski's theorem to prove your statement. $\endgroup$
    – Sergei Golovan
    Mar 1, 2017 at 4:53
  • $\begingroup$ How do we know that once we divide the denominator by $\sqrt{n}\sigma$ that it converges to a constant in order to apply Slutski's Theorem? $\endgroup$
    – user75514
    Mar 1, 2017 at 15:36
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    $\begingroup$ Where does $(1/n)\sum_{i=1}^n\xi_i^2$ converge? $\endgroup$
    – Sergei Golovan
    Mar 1, 2017 at 17:08

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