11. (20 points) Let the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$ be denoted as $\left\{S_{n}\right\}$, and the sum of the first $n$ terms of the sequence $\left\{S_{n}+1\right\}$ be denoted as $\left\{T_{n}\right\}$. Is there a unique sequence $\left\{a_{n}\right\}$ such that the sequence $\left\{T_{n}\right\}$ is the same as the sequence $\left\{S_{n+1}-1\right\}$? If it exists, find the general term formula for the sequence $\left\{a_{n}\right\}$. If it does not exist, find the value of $\lim _{n \rightarrow \infty} \frac{a_{2}+a_{4}+\cdots+a_{2 n}}{a_{1}+a_{3}+\cdots+a_{2 n-1}}$.
11. From the conditions, we know
$$
\left\{\begin{array}{l}
T_{1}=S_{1}+1, \\
T_{2}=(S_{1}+1)+(S_{2}+1)=S_{1}+S_{2}+2, \\
\cdots, \\
T_{n}=\sum_{i=1}^{n}(S_{i}+1)=S_{1}+S_{2}+\cdots+S_{n}+n,
\end{array} \quad \left\{\begin{array}{l}
T_{1}=S_{2}-1 \\
T_{2}=S_{3}-1 \\
\cdots, \\
T_{n}=S_{n+1}-1
\end{array}\right.\right.
$$

Thus, $a_{1}+1=S_{1}+1=T_{1}=S_{2}-1=a_{1}+a_{2}-1 \Rightarrow a_{2}=2, 2a_{1}+a_{2}+2=S_{1}+S_{2}+2=T_{2}=$ $S_{3}-1=a_{1}+a_{2}+a_{3}-1 \Rightarrow a_{3}=a_{1}+3$. For $k \geqslant 2$, we have
$$
\begin{array}{l}
S_{k}-1=T_{k-1}=S_{1}+S_{2}+\cdots+S_{k-1}+(k-1) \\
S_{k+1}-1=T_{k}=S_{1}+S_{2}+\cdots+S_{k}+k .
\end{array}
$$
(2) - (1), we get
$$
\begin{aligned}
a_{k+1} & =\left(S_{k+1}-1\right)-\left(S_{k}-1\right)=\left(S_{1}+S_{2}+\cdots+S_{k}+k\right)-\left(S_{1}+S_{2}+\cdots+S_{k-1}+(k-1)\right) \\
& =S_{k}+1=a_{1}+a_{2}+\cdots+a_{k}+1
\end{aligned}
$$

Thus,
$$
a_{k+2}=a_{1}+a_{2}+\cdots+a_{k}+a_{k+1}+1=2a_{k+1}.
$$

Therefore,
$$
a_{n}=2a_{n-1}=2^{2} \cdot a_{n-2}=\cdots=2^{n-3}a_{3}=2^{n-3}(a_{1}+3) \quad(n \geqslant 3).
$$

Upon verification, for any $t \in \mathbf{R}_{+}$, the sequence
$$
a_{n}=\left\{\begin{array}{ll}
t, & n=1, \\
2, & n=2, \\
2^{n-1}(t+3), & n \geqslant 3
\end{array}\right.
$$

satisfies the conditions. Therefore, the sequence $\left\{a_{n}\right\}$ is not unique. Thus,
$$
\begin{aligned}
\lim _{n \rightarrow \infty} \frac{a_{2}+a_{4}+\cdots+a_{2 n}}{a_{1}+a_{3}+\cdots+a_{2 n-1}} & =\lim _{n \rightarrow \infty} \frac{2+(t+3)\left(2^{1}+2^{3}+\cdots+2^{2 n-3}\right)}{t+(t+3)\left(2^{0}+2^{2}+\cdots+2^{2 n-4}\right)} \\
& =\lim _{n \rightarrow \infty} \frac{\frac{2}{(t+3)\left(2^{0}+2^{2}+\cdots+2^{2 n-4}\right)}+2}{t} \\
& =\lim _{n \rightarrow \infty} \frac{\frac{6}{(t+3)\left(2^{0}+2^{2}+\cdots+2^{2 n-4}\right)}+1}{\frac{6}{(t+3)\left(4^{n-1}-1\right)}+2}=2
\end{aligned}
$$