Let $X = \{ X_{n, m} \}_{n \geq 1, \, 1 \leq m \leq n}$ be a D5163: Random real triangular array such that

(i)	$X_{n, 1}, \ldots, X_{n, n}$ is an D2713: Independent random collection for each $n \in 1, 2, 3, \ldots$
(ii)	$\lambda_1, \lambda_2, \lambda_3, \ldots \in (0, \infty)$ are each a D993: Real number
(iii)	$$\begin{equation} \lim_{n \to \infty} \lambda_n = \infty \end{equation}$$
(iv)	$$\begin{equation} \lim_{n \to \infty} \sum_{m = 1}^n \mathbb{P}(|X_{n, m}| > \lambda_n) = 0 \end{equation}$$
(v)	$$\begin{equation} \lim_{n \to \infty} \frac{1}{\lambda^2_n} \sum_{m = 1}^n \mathbb{E} (|X_{n, m}|^2 I_{\{ |X_{n, m}| \leq \lambda_n \}}) = 0 \end{equation}$$

Fix $\varepsilon > 0$ and denote $Z_{n, m} : = X_{n, m} I_{\{ |X_{n, m}| \leq \lambda_n \}}$, $S^X_n : = \sum_{m = 1}^n X_{n, m}$, $S^Z_n : = \sum_{m = 1}^n Z_{n, m}$, and $\mu_n : = \sum_{m = 1}^n \mathbb{E} Z_{n, m}$. From result R4737: , we have the upper bound $$\begin{equation} \mathbb{P} \left( \left| \frac{S^X_n - \mu_n}{\lambda_n} \right| > \varepsilon \right) \leq \mathbb{P}(S^X_n \neq S^Z_n) + \mathbb{P} \left( \left| \frac{S^Z_n - \mu_n}{\lambda_n} \right| > \varepsilon \right) \end{equation}$$ Notice that, by construction, we have $\{ X_{n, m} \neq Z_{n, m} \} = \{ |X_{n, m}| > \lambda_n \}$ for all $n \in 1, 2, 3, \ldots$. Applying results

(i)	R4740:
(ii)	R4738: Finite subadditivity of probability measure

as well as hypothesis (iv), we can estimate the first term on the right-hand side by $$\begin{equation} \begin{split} \mathbb{P}(S^X_n \neq S^Z_n) \leq \mathbb{P} \left( \bigcup_{m = 1}^n \{ X_{n, m} \neq Z_{n, m} \} \right) & \leq \mathbb{P} \left( \bigcup_{m = 1}^n \{ |X_{n, m}| > \lambda_n \} \right) \\ & \leq \sum_{m = 1}^n \mathbb{P}(|X_{n, m}| > \lambda_n) \\ & \to 0 \end{split} \end{equation}$$ as $n \to \infty$. Next, applying results

(i)	R4741: Probabilistic Chebyshov's inequality for square function
(ii)	R4687: Additivity of variance for a finite number of independent random real numbers
(iii)	R4688: Second moment upper bound to real variance

as well as hypothesis (v), we have $$\begin{equation} \begin{split} \mathbb{P} \left( \left| \frac{S^Z_n - \mu_n}{\lambda_n} \right| > \varepsilon \right) & \leq \frac{1}{\varepsilon^2} \mathbb{E} \left| \frac{S^Z_n - \mu_n}{\lambda_n} \right|^2 \\ & = \frac{1}{\varepsilon^2 \lambda^2_n} \mathsf{Var}(S^Z_n) \\ & = \frac{1}{\varepsilon^2 \lambda^2_n} \sum_{m = 1}^n \mathsf{Var}(Z_{n, m}) \leq \frac{1}{\varepsilon^2 \lambda^2_n} \sum_{m = 1}^n \mathbb{E} Z^2_{n, m} \to 0 \end{split} \end{equation}$$ as $n \to \infty$. Combining these results, we find that $$\begin{equation} \begin{split} \mathbb{P} \left( \left| \sum_{m = 1}^n \frac{X_{n, m} - \mathbb{E} (X_{n, m} I_{\{ |X_{n, m}| \leq \lambda_n \}})}{\lambda_n} \right| > \varepsilon \right) & = \mathbb{P} \left( \left| \sum_{m = 1}^n \frac{X_{n, m} - \mathbb{E} Z_{n, m}}{\lambda_n} \right| > \varepsilon \right) \\ & = \mathbb{P} \left( \left| \frac{\sum_{m = 1}^n X_{n, m} - \sum_{m = 1}^n \mathbb{E} Z_{n, m}}{\lambda_n} \right| > \varepsilon \right) \\ & = \mathbb{P} \left( \left| \frac{S^X_n - \mu_n}{\lambda_n} \right| > \varepsilon \right) \\ & \leq \mathbb{P}(S^X_n \neq S^Z_n) + \mathbb{P} \left( \left| \frac{S^Z_n - \mu_n}{\lambda_n} \right| > \varepsilon \right) \\ & \to 0 \end{split} \end{equation}$$ as $n \to \infty$. $\square$