[proofplan] We prove convergence in distribution by proving pointwise convergence of characteristic functions. First the Lindeberg condition implies that the array is infinitesimal, so no single variance term contributes in the limit. Then a second-order Taylor expansion of $e^{itx}$, with the remainder split into small and large values of $Y_{n,j}$, shows that each row has the same limiting characteristic function as a product of Gaussian characteristic functions with variances $\sigma_{n,j}^2$. The variance convergence identifies the limiting characteristic function as $t \mapsto e^{-\sigma^2 t^2/2}$, and Levy's [continuity theorem](/theorems/1145) gives the desired convergence in distribution. [/proofplan]

[step:Define the row sums and their characteristic functions] For each $n \in \mathbb{N}$, define the row sum [random variable](/page/Random%20Variable) $S_n: (\Omega,\mathcal{F},\mathbb{P}) \to \mathbb{R}$ by \begin{align*} S_n := \sum_{j=1}^{k_n} Y_{n,j}. \end{align*} For each fixed $t \in \mathbb{R}$, define the characteristic function $\varphi_n: \mathbb{R} \to \mathbb{C}$ of $S_n$ by \begin{align*} \varphi_n(t) := \mathbb{E}[e^{itS_n}]. \end{align*} For $1 \leq j \leq k_n$, define the individual characteristic function $\varphi_{n,j}: \mathbb{R} \to \mathbb{C}$ by \begin{align*} \varphi_{n,j}(t) := \mathbb{E}[e^{itY_{n,j}}]. \end{align*} Since $Y_{n,1},\dots,Y_{n,k_n}$ are independent for each fixed $n$, the random variables $e^{itY_{n,1}},\dots,e^{itY_{n,k_n}}$ are independent and bounded. Therefore \begin{align*} \varphi_n(t) = \prod_{j=1}^{k_n} \varphi_{n,j}(t). \end{align*} [/step]

[step:Derive infinitesimality from the Lindeberg condition]We prove that \begin{align*} \max_{1 \leq j \leq k_n} \sigma_{n,j}^2 \to 0. \end{align*} Fix $\varepsilon > 0$. For every $n$ and every $1 \leq j \leq k_n$, split the second moment according to the events $\{|Y_{n,j}| \leq \varepsilon\}$ and $\{|Y_{n,j}| > \varepsilon\}$. Since $Y_{n,j}^2 \leq \varepsilon^2$ on $\{|Y_{n,j}| \leq \varepsilon\}$, we have \begin{align*} \sigma_{n,j}^2 \leq \varepsilon^2 + \mathbb{E}\left[Y_{n,j}^2 \mathbb{1}_{\{|Y_{n,j}|>\varepsilon\}}\right]. \end{align*} Taking the maximum over $j$ and bounding the maximum by the sum gives \begin{align*} \max_{1 \leq j \leq k_n} \sigma_{n,j}^2 \leq \varepsilon^2 + \sum_{j=1}^{k_n}\mathbb{E}\left[Y_{n,j}^2 \mathbb{1}_{\{|Y_{n,j}|>\varepsilon\}}\right]. \end{align*} The Lindeberg condition implies \begin{align*} \limsup_{n \to \infty} \max_{1 \leq j \leq k_n} \sigma_{n,j}^2 \leq \varepsilon^2. \end{align*} Since $\varepsilon > 0$ was arbitrary, the maximum variance tends to $0$.[/step]

[guided]The point of this step is to rule out the possibility that one summand has a macroscopic variance. Fix $\varepsilon > 0$. For each $Y_{n,j}$, we split its second moment into the part where $|Y_{n,j}|$ is at most $\varepsilon$ and the part where it is larger than $\varepsilon$: \begin{align*} \sigma_{n,j}^2 = \mathbb{E}\left[Y_{n,j}^2 \mathbb{1}_{\{|Y_{n,j}|\leq\varepsilon\}}\right] + \mathbb{E}\left[Y_{n,j}^2 \mathbb{1}_{\{|Y_{n,j}|>\varepsilon\}}\right]. \end{align*} On the event $\{|Y_{n,j}| \leq \varepsilon\}$, the inequality $Y_{n,j}^2 \leq \varepsilon^2$ holds. Since $\mathbb{P}(\{|Y_{n,j}| \leq \varepsilon\}) \leq 1$, this gives \begin{align*} \mathbb{E}\left[Y_{n,j}^2 \mathbb{1}_{\{|Y_{n,j}|\leq\varepsilon\}}\right] \leq \varepsilon^2. \end{align*} Therefore \begin{align*} \sigma_{n,j}^2 \leq \varepsilon^2 + \mathbb{E}\left[Y_{n,j}^2 \mathbb{1}_{\{|Y_{n,j}|>\varepsilon\}}\right]. \end{align*} Taking the maximum in $j$ and then bounding that maximum by the whole row sum yields \begin{align*} \max_{1 \leq j \leq k_n} \sigma_{n,j}^2 \leq \varepsilon^2 + \sum_{j=1}^{k_n}\mathbb{E}\left[Y_{n,j}^2 \mathbb{1}_{\{|Y_{n,j}|>\varepsilon\}}\right]. \end{align*} The second term tends to $0$ by the Lindeberg condition. Hence \begin{align*} \limsup_{n \to \infty} \max_{1 \leq j \leq k_n} \sigma_{n,j}^2 \leq \varepsilon^2. \end{align*} Because this is true for every $\varepsilon > 0$, the only possible limit superior is $0$. Thus \begin{align*} \max_{1 \leq j \leq k_n} \sigma_{n,j}^2 \to 0. \end{align*}[/guided]

[step:Control the Taylor remainder row by row] Fix $t \in \mathbb{R}$. Define the Taylor remainder function $r_t: \mathbb{R} \to \mathbb{C}$ by \begin{align*} r_t(x) := e^{itx} - 1 - itx + \frac{t^2x^2}{2}. \end{align*} Since $e^{iu} = 1 + iu - u^2/2 + o(u^2)$ as $u \to 0$, there exists a function $\rho_t: (0,\infty) \to [0,\infty)$ with $\rho_t(\varepsilon) \to 0$ as $\varepsilon \downarrow 0$ such that \begin{align*} |r_t(x)| \leq \rho_t(\varepsilon)x^2 \end{align*} whenever $|x| \leq \varepsilon$. For $|x| > \varepsilon$, use $|e^{itx}|=1$ and the triangle inequality: \begin{align*} |r_t(x)| \leq 2 + |t||x| + \frac{t^2x^2}{2}. \end{align*} Since $|x| > \varepsilon$, we have $1 \leq x^2/\varepsilon^2$ and $|x| \leq x^2/\varepsilon$. Hence \begin{align*} |r_t(x)| \leq C_t(\varepsilon)x^2 \end{align*} on $\{|x|>\varepsilon\}$, where $C_t: (0,\infty) \to (0,\infty)$ denotes the function \begin{align*} \delta \mapsto \frac{2}{\delta^2} + \frac{|t|}{\delta} + \frac{t^2}{2}. \end{align*} For $1 \leq j \leq k_n$, centeredness gives $\mathbb{E}[Y_{n,j}] = 0$, and square-integrability justifies the following expectation: \begin{align*} \varphi_{n,j}(t) = 1 - \frac{t^2\sigma_{n,j}^2}{2} + \mathbb{E}[r_t(Y_{n,j})]. \end{align*} Therefore \begin{align*} \sum_{j=1}^{k_n} \left|\mathbb{E}[r_t(Y_{n,j})]\right| \leq \rho_t(\varepsilon)\sum_{j=1}^{k_n}\sigma_{n,j}^2 + C_t(\varepsilon)\sum_{j=1}^{k_n}\mathbb{E}\left[Y_{n,j}^2\mathbb{1}_{\{|Y_{n,j}|>\varepsilon\}}\right]. \end{align*} Taking $\limsup_{n \to \infty}$ and using the variance convergence and the Lindeberg condition gives \begin{align*} \limsup_{n \to \infty}\sum_{j=1}^{k_n} \left|\mathbb{E}[r_t(Y_{n,j})]\right| \leq \rho_t(\varepsilon)\sigma^2. \end{align*} Letting $\varepsilon \downarrow 0$ gives \begin{align*} \sum_{j=1}^{k_n} \left|\mathbb{E}[r_t(Y_{n,j})]\right| \to 0. \end{align*} [/step]

[step:Compare the characteristic function with the matching Gaussian product] For $1 \leq j \leq k_n$, define the Gaussian comparison factor $g_{n,j}: \mathbb{R} \to \mathbb{C}$ by \begin{align*} g_{n,j}(t) := \exp\left(-\frac{t^2\sigma_{n,j}^2}{2}\right). \end{align*} The elementary inequality $|e^{-a} - (1-a)| \leq a^2/2$ for $a \geq 0$ gives \begin{align*} \left|g_{n,j}(t) - \left(1-\frac{t^2\sigma_{n,j}^2}{2}\right)\right| \leq \frac{t^4\sigma_{n,j}^4}{8}. \end{align*} Combining this with the Taylor expansion from the previous step yields \begin{align*} \sum_{j=1}^{k_n} |\varphi_{n,j}(t)-g_{n,j}(t)| \leq \sum_{j=1}^{k_n}\left|\mathbb{E}[r_t(Y_{n,j})]\right| + \frac{t^4}{8}\sum_{j=1}^{k_n}\sigma_{n,j}^4. \end{align*} The first sum tends to $0$. For the second sum, infinitesimality and the boundedness of the total variance give \begin{align*} \sum_{j=1}^{k_n}\sigma_{n,j}^4 \leq \left(\max_{1 \leq j \leq k_n}\sigma_{n,j}^2\right)\sum_{j=1}^{k_n}\sigma_{n,j}^2 \to 0. \end{align*} Thus \begin{align*} \sum_{j=1}^{k_n} |\varphi_{n,j}(t)-g_{n,j}(t)| \to 0. \end{align*} Since $|\varphi_{n,j}(t)| \leq 1$ and $|g_{n,j}(t)| \leq 1$, the telescoping product inequality gives \begin{align*} \left|\prod_{j=1}^{k_n}\varphi_{n,j}(t)-\prod_{j=1}^{k_n}g_{n,j}(t)\right| \leq \sum_{j=1}^{k_n} |\varphi_{n,j}(t)-g_{n,j}(t)|. \end{align*} Consequently, \begin{align*} \varphi_n(t) - \prod_{j=1}^{k_n}g_{n,j}(t) \to 0. \end{align*} [/step]

[step:Identify the Gaussian limiting characteristic function] By the definition of $g_{n,j}$, \begin{align*} \prod_{j=1}^{k_n}g_{n,j}(t) = \exp\left(-\frac{t^2}{2}\sum_{j=1}^{k_n}\sigma_{n,j}^2\right). \end{align*} Since $\sum_{j=1}^{k_n}\sigma_{n,j}^2 \to \sigma^2$, continuity of the exponential function gives \begin{align*} \prod_{j=1}^{k_n}g_{n,j}(t) \to \exp\left(-\frac{\sigma^2t^2}{2}\right). \end{align*} Combining this with the previous step, for every $t \in \mathbb{R}$, \begin{align*} \varphi_n(t) \to \exp\left(-\frac{\sigma^2t^2}{2}\right). \end{align*} The function $t \mapsto \exp(-\sigma^2t^2/2)$ is the characteristic function of the Gaussian distribution $\mathcal{N}(0,\sigma^2)$. [/step]

[step:Apply Levy's continuity theorem to conclude convergence in distribution] The limiting function $t \mapsto \exp(-\sigma^2t^2/2)$ is continuous at $0$ and is the characteristic function of $\mathcal{N}(0,\sigma^2)$. By Levy's continuity theorem, pointwise convergence of characteristic functions to this limiting characteristic function implies \begin{align*} S_n \xrightarrow{d} \mathcal{N}(0,\sigma^2). \end{align*} Since $S_n = \sum_{j=1}^{k_n}Y_{n,j}$, this is precisely \begin{align*} \sum_{j=1}^{k_n}Y_{n,j} \xrightarrow{d} \mathcal{N}(0,\sigma^2). \end{align*} [/step]