Details on proof of Martingale difference CLT Let $X_1,X_2,\ldots$ be a martingale and $Y_n$ be the martingale differences, which satisfy
\begin{align}
\mathbb{E}[Y_n^2\mid\mathcal{F}_{n-1}]=1,n=1,2,\ldots,\sup_{n\geq 1}\mathbb{E}[|Y_n|^3\mid\mathcal{F}_{n-1}]=K
\end{align}
Let $a_{n,N}=\mathbb{E}(\exp(i \sigma X_n/\sqrt{N}))$.  Then I want to show that
\begin{align}
\left|a_{n,N}-[1-\sigma^2/2N]a_{n-1,N}\right|&\leq K\sigma^3 /6N^{3/2}
\end{align}
I'm looking at equation (2.23) of prop 2.2 on p. 4 of http://www.math.lsa.umich.edu/~conlon/math625/chapter2.pdf.  I can't figure out why it's true.
 A: First observe that $\exp\left(i\sigma\frac{X_{n-1}}{\sqrt N}\right)$ is $\mathcal F_{n-1}$-measurable hence
$$
a_{n,N}=\mathbb E\left[\mathbb E\left[\exp\left(i\sigma\frac{X_n}{\sqrt N}\right)\mid \mathcal F_{n-1}\right]\right]=\mathbb E\left[\exp\left(i\sigma\frac{X_{n-1}}{\sqrt N}\right)\mathbb E\left[\exp\left(i\sigma\frac{Y_n}{\sqrt N}\right)\mid \mathcal F_{n-1}\right]\right].
$$
It follows that 
$$
a_{n,N}-[1-\sigma^2/2N]a_{n-1,N}=\mathbb E\left[\exp\left(i\sigma\frac{X_{n-1}}{\sqrt N}\right)\left(\mathbb E\left[\exp\left(i\sigma\frac{Y_n}{\sqrt N}\right)\mid \mathcal F_{n-1}\right]-[1-\sigma^2/2N]  \right)\right]
$$
and since $\exp\left(i\sigma\frac{X_{n-1}}{\sqrt N}\right)$ is bounded by $1$, we get 
$$
\left|a_{n,N}-[1-\sigma^2/2N]a_{n-1,N}\right|\leqslant\left\lvert \mathbb E\left[\exp\left(i\sigma\frac{Y_n}{\sqrt N}\right)\mid \mathcal F_{n-1}\right]-[1-\sigma^2/2N]\right\rvert. 
$$
Then use Taylor's formula and the assumption on the conditional moments to get the wanted inequality.
