Let $\xi_1, \xi_2...$ be i.i.d. random variables. $\mathbb{E}\xi_i = 0, \mathbb{E}\xi^2_i = 1 \ \forall i $. Let $\lambda_1, \lambda_2, ...$ be a sequence of real number such that
$$ \frac{\max_{k \in \{1, ..., n\}} \lambda_k^2}{\sum\limits_{i=1}^n \lambda_i^2} \longrightarrow 0 $$
Prove that CLT is applicable for sequence of random variables $\lambda_1 \xi_1, \cdots \lambda_2 \xi_n\cdots$.
That's my solution.
Let $\phi(t)$ be a characteristic function of $\xi$. Then remembering, that $\mathbb{E}\xi = 0, \mathbb{E}\xi^2 = 1$, one can see
$$ \phi(t) = 1 - \frac{t^2}{2} + r(t), $$
where $r(t) = o(t^2)$.
Let $\zeta_n$ be equal to $\frac{\lambda_1 \xi_1 + ... + \lambda_n \xi_n}{\sqrt{\lambda_1^2 + ... + \lambda_n^2}}$. Then, using independence of random variable, we get
$$ \phi_{\zeta_n}(t) = \prod_{i=1}^n\phi\left(\frac{\lambda_i}{\sqrt{\lambda_1^2 + ... + \lambda_n^2}}t\right) $$
Taking $\log $ of both sides one gets
$$ \log{\phi_{\zeta_n}(t)} = \sum\limits_{i=1}^n \log{\left(1 - \frac{\lambda_i^2}{\lambda_1^2 + ... + \lambda_n^2}\frac{t^2}{2} + r\left(\frac{\lambda_i}{\sqrt{\lambda_1^2 + ... + \lambda_n^2}}t\right)\right)} $$
Considering Taylor's series, let's recall $\log(1 + x) = \sum\limits_{k=1}^{\infty} \frac{(-1)^{k-1} x^k}{k!}$.
Therefore if $\log(1+x) = x + \alpha(x)$ and $|x|\leq \frac12$, then
$$ |\alpha(x)| = |\sum\limits_{k=2}^{\infty} \frac{(-1)^{k-1} x^k}{k!}| \leq \frac 12 \sum\limits_{k = 2}^{\infty} |x|^k \leq |x|^2. $$
For characteristic functions it can be proved that (considering $\mathbb{E} \xi = 0)$
$$ |\phi_{\xi}(t) - 1| \leq \frac{t^2}{2} \cdot \mathbb{E}\xi^2 = \frac{t^2}{2} $$
Let's fix $T > 0$ and choose $n_0$ so that $\forall n \geq n_0$ $\max \frac{\lambda_i^2}{\lambda_1^2 + \cdots \lambda_n^2} \frac{T^2}{2} \leq \frac 12$. Then $\forall t \in [-T, T]$
$$ |\phi_{\zeta_n} + \frac{t^2}{2}| \leq \sum\limits_{i=1}^n \left|r\left(\frac{\lambda_i}{\sqrt{\lambda_1^2 + ... + \lambda_n^2}}t\right)\right| + \sum\limits_{i=1}^n \left|\frac{\lambda_i^2}{{\lambda_1^2 + ... + \lambda_n^2}} \frac{t^2}{2}\right|^2 \leq $$ $$ \leq \sum\limits_{i=1}^n \left|r\left(\frac{\lambda_i}{\sqrt{\lambda_1^2 + ... + \lambda_n^2}}t\right)\right| + \frac{T^4}{4} \max_i \frac{\lambda_i^2}{\lambda_1^2 + \cdots \lambda_n^2} \cdot \sum\limits_{i=1}^n \frac{\lambda_i^2}{\lambda_1^2 + ... + \lambda_n^2} = $$ $$ = \sum\limits_{i=1}^n \left|r\left(\frac{\lambda_i}{\sqrt{\lambda_1^2 + ... + \lambda_n^2}}t\right)\right| + \frac{T^4}{4} \max_i \frac{\lambda_i^2}{\lambda_1^2 + \cdots \lambda_n^2} $$
Last step is to evaluate first sum. Let's recall that $r(t) = o(t^2)$. So $\forall t_0 \in (-T, T) \ \exists C_{t_0}: \ |r(t)|\leq C_{t_0} \cdot t^2$ on $(-t_0, t_0)$, and $C_{t_0}$ converges to $0$ while $t_0$ converges to $0$.
$t':= \max \frac{\lambda_i}{\sqrt{\lambda_1^2 + \cdots + \lambda_n^2}} T$
Then (it's an only place where I use that all random variables are identically distributed) $$ \sum\limits_{i=1}^n \left|r\left(\frac{\lambda_i}{\sqrt{\lambda_1^2 + ... + \lambda_n^2}}t\right)\right| \leq C_{t'} T^2 \sum\limits_{i=1}^n \left(\frac{\lambda_i}{\sqrt{\lambda_1^2 + ... + \lambda_n^2}}\right)^2 = C_{t'}T^2 $$
Finally, we get $$ \left|\phi_{\zeta_n}(t) + \frac{t^2}{2}\right| \leq C_{t'}T^2 + \frac{T^4}{4} \max_i \frac{\lambda_i^2}{\lambda_1^2 + \cdots \lambda_n^2} $$
The right side converges to $0$ on $[-T, T]$ while $n \longrightarrow \infty$.