Let $E_n=\{X_n\ne a_n\}$ and $E=\cap_{n=1}^\infty \cup_{k=n}^\infty E_k$. Note that if $\omega\notin E$ then we have $X_n(\omega)\ne a_n$ for only finitely many values of $n$, so the sequence $X_n(\omega)$ eventually just becomes the sequence $a_n$, and hence $\sum_{n=1}^\infty X_n(\omega)$ converges. So now we just have to prove that the probability of the event $\{\omega\notin E\}$ is $1$, or equivalently that $\mathbb{P}(E)=0$. But this just follows immediately from the first Borel-Cantelli lemma.
Let $\{X_n\}_{n \in \mathbb N}$ be a family of i.i.d r.vs with distribution $\mu$. Let $A \in \mathcal B(\mathbb R)$ be such that $\mu(A) \in (0,1)$. Define $\tau_A := \inf \{ n \ge 1 : X_n \in A \}$.
a) Note that $\tau_A$ has geometric distribution with parameter $\mu(A)$. Indeed, for $n \ge 1$:
$$ \mathbb P(\tau_A = n) = \mathbb P(X_1,...,X_{n-1} \not \in A, X_n \in A) = \mu(A)(1-\mu(A))^{n-1} $$
So (since $\mu(A) \in (0,1)$ !)
$$ \mathbb P(\tau_A < \infty) = \sum_{n=1}^\infty \mathbb P(\tau_A = n) = \mu(A)\sum_{n=1}^\infty (1-\mu(A))^{n-1} = \frac{\mu(A)}{1-(1-\mu(A))} = 1 $$
b) Let $X_{\tau_A}(\omega) = X_{\tau_A(\omega)}(\omega)$ (it is well defined up to the set of measure $0$)
Let any $H \in \mathcal B(\mathbb R)$. We have:
$$ \mathbb P(X_{\tau_A} \in H) = \sum_{n=1}^\infty \mathbb P(X_n \in H \cap \{ \tau_A = n \}) = \sum_{n=1}^\infty \mathbb P(\{X_1,...,X_{n-1} \not \in A\} \cap \{X_n \in H \cap A \}) $$
Last equality, during to the fact, that since $\tau_A = n$, we need every $X_1,...,X_{n-1}$ not to fall into $A$, and $X_n$ must fall into $A$ and $H$ simultaneously. Since they are independent, we get it is equal to $(1-\mu(A))^{n-1} \cdot \mu(H \cap A)$, so:
$$ \mathbb P(X_{\tau_A} \in H) = \mu(A\cap H) \sum_{n=1}^\infty (1-\mu(A))^{n-1} = \frac{\mu(A \cap H)}{\mu(A)}$$
Best Answer
Define $$ G(t)=\begin{cases}\dfrac{e^t-\mu}{t-\mu}, & t\neq \mu,\cr e^\mu, & t=\mu.\end{cases} $$ This function is continuous everywhere since $\lim\limits_{t\to\mu}\dfrac{e^t-\mu}{t-\mu}=(e^t)'_\mu=e^\mu$. By LLN, $\frac{S_n}{n}\xrightarrow{p}\mu$, and by continuous mapping theorem, $$ G\left(\frac{S_n}{n}\right)\xrightarrow{p}G(\mu)=e^\mu. $$ Then $$ \sqrt{n}\left(e^{\frac{S_n}{n}}-e^{\mu}\right)=\sqrt{n}\left(\frac{S_n}n-\mu\right)\cdot G\left(\frac{S_n}{n}\right). $$ Here the first term $\sqrt{n}\left(\frac{S_n}n-\mu\right)\xrightarrow{d}\sigma Z$ by CLT, where $Z\sim N(0,1)$. And the second term converges in probability to $e^\mu$. Slutsky's theorem implies that the product converges in distribution to $\sigma e^\mu Z$.
Note that this is almost the same reasoning as in the previous answer, and indeed the same as the so-called delta method.