I'm reading the CLT proof and am struggling with the following:

(I'm skipping some details of the proof and only getting to the part I don't understand).

The complex logarithm has $\log{(1+z)}=z+o(|z|)$ as $z\to 0$.

So, for $t\in\mathbb{R}$ fixed, as $n\to\infty$, $$n\log{\left(1-t^2/(2n)+o(t^2/n)\right)}=-t^2/2 + o(1).$$

My question is how exactly does he get this equality. I've tried using $\log{(1+z)}=z+o(|z|)$ from above but I still can't see it. Any help will be appreciated!

  • 1
    $\begingroup$ Use the expansion of $\log(1+z)$ for $z=-t^2/(2n)+o(t^2/n)$ and note that, for every fixed $t$, $n\cdot o(t^2/n)=o(1)$ when $n\to\infty$. $\endgroup$ – Did Sep 18 '15 at 17:26

Personally, when I am using little-o notation (or big-O for that matter) I prefer to write in terms of explicit functions and then convert to little-o at the end. So we write $$\log(1+z)=z+R(z)$$ where $R(z)/|z|\to0$ as $z\to0$. For the $o(t^2/n)$ term, I will similarly write it as $S(t^2/n)$ where $S(z)/|z|\to0$ as $z/to0$. Then $$ n\log(1-t^2/(2n)+S(t^2/n))=n[-t^2/(2n)+S(t^2/n)]+R(t^2/(2n))\\ =-t^2/2+nS(t^2/n)+nR(t^2/(2n)) $$ and clearly both the $R$ and $S$ terms go to zero as $n\to\infty$.


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