Differentiability gives $\varphi(u)=1+iau+o(u)$ at zero. By the <characteristic function of a sum of independent variables>,
$$
\mathbb E e^{itS_n/n}=\varphi(t/n)^n\longrightarrow e^{iat},
$$
the characteristic function of the constant $a$. The <Lévy continuity theorem> yields $S_n/n\xrightarrow d a$, and convergence in distribution to a constant is equivalent to <convergence in probability>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 201 / 5 / c / Solution

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If the <characteristic functions> $\varphi_n$ of <probability distributions> $\mu_n$ converge pointwise to a function $\varphi$ that is continuous at zero, then $\varphi$ is a characteristic function and $\mu_n$ <convergence in distribution>[converges in distribution] to its distribution. Conversely, convergence in distribution implies pointwise convergence of the characteristic functions.

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 Incoming links: Lévy continuity theorem