# A case of the central limit theorem

I want to show that

$$\frac{\sum_{k=1}^N X_k}{\sqrt{\sum_{k=1}^N X_k^2}} \overset{N\to\infty}{\to} \mathcal{N}(0,1)\text{ in distribution,}$$

where $X_1,X_2,\ldots$ is a sequence of iid random variables with $\mathbb{E}(X_1)=0$ and $\mathbb{E}(X_1^2) = s < \infty$.

Now I know about the CLT, i. e. $\frac{\sum_{k=1}^N X_k}{\sqrt{sn}}\to\mathcal{N}(0,1)$, and about the proof with characteristic functions, but calculating the CF of this thing seems a bit cumbersome and I think I am missing a more elegant approach.

I would appreciate a hint. TIA

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Divide numerator and denominator by $\sqrt{sn}$ and apply Slutsky's theorem. –  guy Jun 15 '13 at 23:09
This helped a lot, thank you very much! –  Fye Jun 15 '13 at 23:40
I thought of mentioning Slutsky's theorem; then I saw the comment above. I'll add that all sorts of stuff (including Slutsky's theorem) is in Robert Serfling's book Approximation Theorems of Mathematical Statistics. –  Michael Hardy Jun 15 '13 at 23:42
@guy Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. –  Julian Kuelshammer Jun 25 '13 at 7:11

Divide numerator and denominator by $\sqrt {sn}$ and apply Slutsky's theorem. After the division, the numerator converges to an $\mathcal N(0, 1)$ while $$\sqrt{\sum X_i ^ 2 / sn} \to \sqrt{s / s} = 1,$$ by the SLLN. By Slutsky's theorem the ratio converges to an $\mathcal N(0, 1)$ random variable.