An easy proof for $\frac{S_n}{n}$ converges to $\mu$ in distribution Question:
Consider iid random variables $X_1,X_2,X_3,\dots$ with finite mean $\mu$. Let $S_n = X_1+X_2+\cdots+X_n$. Show that $\frac{S_n}{n}$ converges to $\mu$ in distribution as $n\rightarrow \infty$.

Attempt:
So this is basically a weaker version of the Weak Law of Large Numbers. Of course, I could do this by first proving the WLLN (which requires proof of Markov and Chebyshev inequalities), then prove that convergence in probability implies convergence in distribution.
But is there a more direct way of doing this?
 A: This answer is essentially the same as @Kavi Rama Murthy's one, with the additional step of establishing the desired error term.
Let $(X_n)_{n\geq 1}$ be i.i.d. and $\mathbb{E}[|X_n|] < \infty$. Using the identity $ e^{z} = 1 + z \int_{0}^{1} e^{zs} \, \mathrm{d}s$ which holds for any $z \in \mathbb{C}$, we find that
$$ \mathbb{E}\big[e^{it X_1/n}\big]
= \mathbb{E}\bigg[ 1 + \frac{itX_1}{n} \int_{0}^{1} e^{istX_1/n} \, \mathrm{d}s \bigg]
= 1 + \frac{it}{n} \mathbb{E}\bigg[ \int_{0}^{1} X_1 e^{istX_1/n} \, \mathrm{d}s \bigg] $$
for any $t \in \mathbb{R}$ and $n \geq 1$. Also, by the Dominated Convergence Theorem, we get
$$ \mathbb{E}\bigg[ \int_{0}^{1} X_1 e^{istX_1/n} \, \mathrm{d}s \bigg] \xrightarrow[n\to\infty]{} \mathbb{E}\bigg[ \int_{0}^{1} X_1 \, \mathrm{d}s \bigg] = \mu $$
where $\mu = \mathbb{E}[X_1]$. So it follows that
$$ \mathbb{E}\big[ e^{itS_n/n} \big] = \mathbb{E}\big[ e^{itX_1/n} \big]^n = \left( 1 + \frac{it\mu +o(1)}{n}\right)^n \xrightarrow[n\to\infty]{}e^{it\mu} $$
for any $t \in \mathbb{R}$, and therefore $S_n/n \to \mu$ in distribution by the Lévy's Continuity Theorem.
