Try $\Omega=\{0,1\}^{\mathbb N}$, $\mathcal F_n=\{A\times\{0,1\}^{\{n+1,n+2,\ldots\}}\mid A\subseteq\{0,1\}^n\}$ and $X_n(\omega)=\omega_n$ for every $n\geqslant0$.
Thus, $\Omega$ is the set of all infinite $\{0,1\}$-valued sequences, not the union on $n\geqslant0$ of the sets of $\{0,1\}$-valued sequences of length $n$.
Edit: An alternative characterization of $\mathcal F_n$ is that $B\subseteq\Omega$ is in $\mathcal F_n$ if and only if property $P_n(B)$ holds:
$P_n(B):$ Let $\omega=(\omega_k)_{k\in\mathbb N}$ and $\omega'=(\omega'_k)_{k\in\mathbb N}$ denote two elements of $\Omega$. If $\omega$ is in $B$ and if $\omega'_k=\omega_k$ for every $1\leqslant k\leqslant n$, then $\omega'$ is in $B$.
Note that each $\mathcal F_n$ contains $\varnothing$ since $\varnothing=\varnothing\times\{0,1\}^{\{n+1,n+2,\ldots\}}$ and $\varnothing\subseteq\{0,1\}^n$, and that each $\mathcal F_n$ contains $\Omega$ since $\Omega=\{0,1\}^n\times\{0,1\}^{\{n+1,n+2,\ldots\}}$ and $\{0,1\}^n\subseteq\{0,1\}^n$. Furthermore, for each $n$, $B_n=\{0\}^n\times\{0,1\}^{\{n+1,n+2,\ldots\}}$ is in $\mathcal F_k$ if and only if $n\leqslant k$. An equivalent formulation of the set $B_n$ is that $B_n=\bigcap\limits_{i=1}^n[X_i=0]$.
Edit-edit: The above corresponds to the description of the sequence $(\mathcal F_n)_n$ in the post. However, in the context of tail sigma-algebras, a more natural choice would be $\mathcal F_n$ the sigma-alebra generated by $X_n$ alone (not by every $X_k$ with $k\leqslant n$ as above). Then $B\subseteq\Omega$ is in $\mathcal F_n$ if and only if property $Q_n(B)$ holds:
$Q_n(B):$ Let $\omega=(\omega_k)_{k\in\mathbb N}$ and $\omega'=(\omega'_k)_{k\in\mathbb N}$ denote two elements of $\Omega$. If $\omega$ is in $B$ and if $\omega'_n=\omega_n$, then $\omega'$ is in $B$.
Then the tail sigma-algebra $\mathcal T=\bigcap\limits_n\bigcup\limits_{k\geqslant n}\mathcal F_k$ contains events such as $[X_n=1\ \text{for infinitely many}\ n]$ or $[S_n/n\to c]$ for every $c$ (the proofs are not that difficult but this should be proven), and a powerful result in this domain is the so-called Kolmogorov zero-one law, which states that, if the sequence $(\mathcal F_n)$ is independent then $\mathcal T$ contains events of probability $0$ or $1$ only.
The answer follows the idea of what @d.k.o. pointed out. For example, for the statement
$$\{ \omega: \exists \lim_{n \to \infty} X_n(\omega) \} \in C_\infty$$
we can argue as follows:
For each n, $X_{m}$ with $m \ge n$ is measurable for the $\sigma$-field ${C}_n
= \sigma(\cup_{p \ge n} B_{p})$. Now we use the following fact:
The set of convergence of a sequence of measurable functions is measurable
Therefore the set of convergence of the sequence ${X_{m}}$ with $m \ge n$ is measurable in $C_{n}$. As this set of convergence does not change with n, it follows it belongs to $C_\infty$.
When you understand this proof the rest is easier. Let's prove that:
$\limsup_{n \to \infty} X_n$ is a measurable function in $C_\infty$.
$[\lim sup_{n \to \infty} X_{n} < c]=[\lim sup_{k \to \infty, k \ge n} X_{n} < c]$ which belongs to $C_{n}$ and the statement follows.
Best Answer
For any $k$. $\lim S_n$ exists iff $\lim_n (X_k+X_{k+1}+...+X_n)$ exists. Applying your expression for the event $(\lim S_n)$ exists for the sequence $X_k,X_{k+1},...$ we see that the event '$\lim_n (X_k+X_{k+1}+...+X_n)$ exists' belongs to $\sigma (X_k,X_{k+1},...)$. Since this is true for each $k$ it follows that the event '$\lim S_n$ exists ' belongs to the tail sigma field. The argument for the other two cases are similar. For the third case you have to observe that $S_k /c_n \to 0$ for each $k$ so the given event does not change if you replace $(X_k)$ by the sequence $(X_k,X_{k+1},...)$.