I am having some trouble understanding a couple of lines in the proof of the central limit theorem using characteristic functions:


I'm not sure how $(1-\frac{t^2}{2n} + o(\frac{t^2}{n}))^n \to e^\frac{-t^2}{2}$ as $n \to \infty$.
Originally I proved (by induction) a lemma that for complex numbers $w_k$ and $z_k$ if $|w_k|\leq 1$, $|z_k|\leq 1$ then $$|\prod_k{w_k}-\prod_k {z_k}| \le \sum_k{|z_k-w_k|}$$ I tried to use this to show that $(1-\frac{t^2}{2n} + o(\frac{t^2}{n}))^n - (1-\frac{t^2}{2n})^n \to 0$ as $n\to \infty$. But I got stuck.

Instead, I've taken the logarithm of the characteristic function (see the link) and then $$n\ln[1-\frac{-t^2}{2n} + o(\frac{t^2}{n})] \stackrel{?}{=} n[\frac{-t^2}{2n} + o(\frac{t^2}{n})] = \frac{-t^2}{2} + o(1) \to \frac{-t^2}{2}$$

However, I'm not sure how to get the first equality, indicated by $?$

Any help would be greatly appreciated.

Thanks in advance.

  • 1
    $\begingroup$ Let $\epsilon>0$. Consider $$(1-\frac{t^2}{2n}- \epsilon \frac{t^2}{2n})^n \rightarrow e^{(-1-\epsilon)t^2/2}$$ and $$(1-\frac{t^2}{2n}+ \epsilon \frac{t^2}{2n})^n \rightarrow e^{(-1+\epsilon)t^2/2}.$$ $\endgroup$ – Sungjin Kim Feb 15 at 4:00
  • 1
    $\begingroup$ So your using the squeeze theorem? $\endgroup$ – Ditherer Feb 15 at 4:28

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