Stuck on proving a variation of the Central Limit Theorem. I'm trying to prove the following version of the central limit theorem.
Let $$L^n = (L_1^n,...L_n^n) $$
such that the $L_i$ are i.i.d.,
there exists a sequence of constants such that $|L_i^n|\leq K^n$ for all $i$ and $K^n \rightarrow{0}$,
and it holds that for $Z_n= \sum_{i=1}^nL_i^n$ we have $\mathbb{E}[Z_n] \rightarrow{\mu}$ and $\operatorname{Var}(Z_n) \rightarrow \sigma^2$.
Then we have that $Z_n$ converges in distribution to $Z$ where $Z$ is normal with mean $\mu$ and variance $\sigma^2$.
I think it should be done with Characteristic functions + Levy's continuity theorem. The characterstic function will factor by the first hypothesis but I don't see how to incorporate the other two to arrive at the desired result.
 A: Yes, you are right, the argument uses Levy's Convergence theorem and is like the proof for CLT in the iid case.
Let $\mu_n = \mathbb{E}\left[L^n_1\right]$ and $\sigma_n^2 = \text{Var}\left(L^n_1\right)$, then by the assumptions on $Z$,
$$\mathbb{E}\left[Z_n\right] = n \mu_n \to \mu \quad \text{ and } \quad \text{Var}\left(Z_n\right) = n \sigma_n \to \sigma, \qquad (*)$$
which implies that $\mu_n = O\left(\frac{1}{n}\right)$ and $\sigma^2_n = O\left(\frac{1}{n}\right)$.
In a similar spirit to estimates derived in the proof of the standard CLT (see for example section 18.3 of D. Williams's Probability with Martingales), let $R_2(x) = e^{ix} - \left(1 + ix - \frac{x^2}{2}\right)$ be the remainder of second order Taylor approximation of $e^{ix}$. Additionally, it holds that $\left|R_2(x)\right| \leq \frac{|x|^3}{6}$, see the Williams's book if it isn't clear to you why this is true. Then we have for the characteristic function of $L_1^n - \mu_n$, the following estimate,
$$\varphi_{L_1^n - \mu_n}\left(\theta\right) = \mathbb{E}\left[\exp{\left(i\theta \left(L_1^n - \mu_n\right)\right)}\right] = 1 - \frac{\theta^2\sigma_n^2}{2} + \mathbb{E}\left[R_2\left(\theta L_1^n\right)\right] = 1 - \frac{\theta^2\sigma_n^2}{2} + R^n\left(\theta\right),$$
where $\left|R^n\left(\theta\right)\right| \leq \frac{\left|\theta\right|^3}{6}\mathbb{E}\left[\left|L_1^n - \mu_n\right|^3\right]$. Applying Hölder and the bound on $L_1^n$, we have
$$\left|R^n\left(\theta\right)\right| \leq \left\|L_1^n - \mu_n\right\|_\infty\mathbb{E}\left[\left(L_1^n - \mu_n\right)^2\right]\leq \frac{\left|\theta\right|^3}{6}\left(K_n + \left|\mu_n\right|\right)\mathbb{E}\left[\left(L_1^n - \mu_n\right)^2\right] = \frac{\left|\theta\right|^3}{6}\left(K_n + \left|\mu_n\right|\right) \sigma_n^2,$$
here $\|\cdot\|_\infty$ is the sup-norm.
The above implies that $R^n = o\left(\frac{1}{n}\right)$ as $K_n, \left|\mu_n\right| \to 0$ as $n\to \infty$ and $\sigma^2_n = O\left(\frac{1}{n}\right)$.
Denote by $\varphi_{Z_n}$ the characteristic function of $Z_n$, which is given by
$$\varphi_{Z_n}\left(\theta\right) = \mathbb{E}\left[\exp{\left(i\theta Z_n\right)}\right] = \prod_{i = 1}^n\varphi_{L_i^n} =  e^{i\theta \mu_n n}\left(\varphi_{L_1^n - \mu_n}\left(\theta\right)\right)^n. $$
Taking logs and applying the asymptotics we derived and $(*)$, we get
\begin{gather}
\log{\varphi_{Z_n}\left(\theta\right)}  
= i\theta \mu_n n + n\log{\left(1 - \frac{\theta^2 \sigma_n^2}{2} + R_n\left(\theta\right)\right)} \\
= i\theta \mu_n n - \frac{\theta^2\sigma_n^2 n}{2} + n R_n\left(\theta\right) + O\left(\frac{1}{n}\right) \xrightarrow{n \to \infty} i\theta \mu - \frac{\theta^2\sigma^2}{2},
\end{gather}
as $nR_n\left(\theta\right) \to 0$. Finally Levy's Convergence theorem lets us conclude that $Z_n$ converges weakly to $Z \sim \mathcal{N}\left(\mu, \sigma\right)$.
