[proofplan] We compute the characteristic function of $S_n/\sqrt{n}$, show it converges pointwise to the Gaussian characteristic function $e^{-u^2/2}$, and apply the [Levy Continuity Theorem](/theorems/519). The key ingredients are: independence (to factor the joint characteristic function), the Taylor expansion $\phi(t) = 1 - t^2/2 + o(t^2)$ (from the moment assumptions), and the classical limit $(1 + a/n + o(1/n))^n \to e^a$. [/proofplan] [step:Factor the characteristic function of $S_n/\sqrt{n}$ using independence] Let $\phi(u) = \mathbb{E}[e^{iuX_1}]$ be the common characteristic function. By independence, the characteristic function of $S_n = X_1 + \cdots + X_n$ factors as a product. Evaluating at $u/\sqrt{n}$: \begin{align*} \mathbb{E}\!\left[e^{iu S_n/\sqrt{n}}\right] = \prod_{k=1}^n \mathbb{E}\!\left[e^{iu X_k/\sqrt{n}}\right] = \phi\!\left(\frac{u}{\sqrt{n}}\right)^n. \end{align*} [/step] [step:Establish the second-order Taylor expansion of $\phi$] [claim:Second Order Expansion Of Characteristic Function] Under the hypotheses $\mathbb{E}[X_1] = 0$ and $\mathbb{E}[X_1^2] = 1$, the characteristic function satisfies \begin{align*} \phi(t) = 1 - \frac{t^2}{2} + o(t^2) \quad \text{as } t \to 0. \end{align*} [/claim] [proof] Differentiating under the integral sign (justified by dominated convergence, since $|ixe^{itx}| = |x|$ and $|{-x^2 e^{itx}}| = x^2$, both integrable): \begin{align*} \phi'(t) = i\int_{\mathbb{R}} x\,e^{itx}\,\mu_{X_1}(dx), \quad \phi''(t) = -\int_{\mathbb{R}} x^2\,e^{itx}\,\mu_{X_1}(dx). \end{align*} Evaluating: $\phi(0) = 1$, $\phi'(0) = i\,\mathbb{E}[X_1] = 0$, $\phi''(0) = -\mathbb{E}[X_1^2] = -1$. By Taylor's theorem with integral remainder, \begin{align*} \phi(t) = 1 - \frac{t^2}{2} + r(t), \end{align*} where $|r(t)| \leq \frac{t^2}{2}\sup_{|s| \leq |t|} |\phi''(s) - \phi''(0)|$. The function $\phi''$ is continuous at $0$ by the Dominated Convergence Theorem (applied to $x^2 e^{isx} \to x^2$ as $s \to 0$ with dominator $x^2 \in L^1$), so $r(t) = o(t^2)$. [/proof] [/step] [step:Show $\phi(u/\sqrt{n})^n \to e^{-u^2/2}$ using the logarithmic limit] [claim:Power Convergence To Gaussian] For every $u \in \mathbb{R}$, \begin{align*} \phi\!\left(\frac{u}{\sqrt{n}}\right)^n \to e^{-u^2/2} \quad \text{as } n \to \infty. \end{align*} [/claim] [proof] Write $t = u/\sqrt{n}$, so $\phi(t) = 1 - u^2/(2n) + o(u^2/n)$ by the previous step (since $t^2 = u^2/n$). Taking logarithms: \begin{align*} n\log\phi\!\left(\frac{u}{\sqrt{n}}\right) = n\log\!\left(1 - \frac{u^2}{2n} + o\!\left(\frac{u^2}{n}\right)\right). \end{align*} Using $\log(1 + z) = z + O(z^2)$ for $|z|$ small, with $z = -u^2/(2n) + o(u^2/n)$: \begin{align*} n\log\phi\!\left(\frac{u}{\sqrt{n}}\right) = n\!\left(-\frac{u^2}{2n} + o\!\left(\frac{u^2}{n}\right) + O\!\left(\frac{u^4}{n^2}\right)\right) = -\frac{u^2}{2} + o(1), \end{align*} so $\phi(u/\sqrt{n})^n = \exp(n\log\phi(u/\sqrt{n})) \to e^{-u^2/2}$. [/proof] [/step] [step:Apply the Levy Continuity Theorem to conclude convergence in distribution] The function $u \mapsto e^{-u^2/2}$ is the characteristic function of the standard normal distribution $N(0,1)$. By the [Levy Continuity Theorem](/theorems/519), the pointwise convergence $\phi(u/\sqrt{n})^n \to e^{-u^2/2}$ for all $u \in \mathbb{R}$ implies $S_n/\sqrt{n} \xrightarrow{d} N(0,1)$. [/step]