Can anyone explain to me where the circled inequality comes from in the following proof of the central limit theorem?
It says that Lemma 2 is used to obtain this, so I have attached a screenshot of this Lemma also:
I have tried to obtain this equality numerous times, but have been unable to do so successfully. Could someone please explain how it has arisen?
EDIT (some attempted workings): I have deduced thus far that, using the substitution $$ \xi = \frac{\xi_1 - a}{\sigma \sqrt{n}} $$ that $$ \mathbb{E} \left[ \xi \right] = \frac{\mathbb{E} \left[ \xi_1 \right]}{\sigma \sqrt{n}} - \frac{a}{\sigma \sqrt{n}} = 0 \\ \mathbb{E} \left[ \xi^2 \right] = \frac{\mathbb{E} \left[ \xi_1^2 \right]}{\sigma^2 n} - \frac{\mathbb{E} \left[ \xi_1 \right]}{\sigma^2 n} + \frac{a}{\sigma^2 n} = \frac{b^2}{\sigma^2 n} $$
Substituting this into the formula given in theorem 2, we get $$ \mathbb{E} \left[ e^{i \lambda \xi} \right] = 1 - \frac{b \lambda^2}{2 \sigma^2 n} + o(\lambda^2) = 1 - \frac{\sigma^2 \lambda^2}{2 \sigma^2 n} + o(\lambda^2) = 1 - \frac{\lambda^2}{2 n} + o(\lambda^2) $$
So I see now that this was relatively easily done?
To clarify though, is the $o(\lambda^2)$ term just some arbitrary function such that $o(\lambda^2) \rightarrow 0$ faster than $\lambda^2 \rightarrow 0$?