11. (20 points) Given the sequence $\left\{a_{n}\right\}$:
$1,1,2,1,2,3, \cdots, 1,2, \cdots, n, \cdots$
Let $S_{n}$ be the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$. Find all positive real number pairs $(\alpha, \beta)$ such that
$$
\lim _{n \rightarrow+\infty} \frac{S_{n}}{n^{\alpha}}=\beta .
$$
11. From the problem, we know that the integer $n$ first appears in the sequence $\left\{a_{n}\right\}$ at the position
$$
1+2+\cdots+n=\frac{n(n+1)}{2} .
$$

Let $A_{n}=\frac{S_{n}}{n^{\alpha}}, B_{n}=\frac{S_{\frac{n(n+1)}{2}}^{2}}{\left[\frac{n(n+1)}{2}\right]^{\alpha}}(n=1,2, \cdots)$.
Then $B_{n}=\frac{\sum_{k=1}^{\frac{n(n+1)}{2}} a_{k}}{\left[\frac{n(n+1)}{2}\right]^{\alpha}}$
$$
\begin{array}{l}
=\frac{\sum_{k=1}^{n}(1+2+\cdots+k)}{\left[\frac{n(n+1)}{2}\right]^{\alpha}}=\frac{\sum_{k=1}^{n} \frac{k(k+1)}{2}}{\left[\frac{n(n+1)}{2}\right]^{\alpha}} \\
=\frac{\frac{1}{6} n^{3}\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)}{\left(\frac{1}{2}\right)^{\alpha} n^{2 \alpha}\left(1+\frac{1}{n}\right)^{\alpha}} .
\end{array}
$$

When $\alpha\frac{3}{2}$, the sequence $\left\{B_{n}\right\}$ converges to 0.
Thus, $\alpha=\frac{3}{2}$. At this time, $\lim _{n \rightarrow+\infty} B_{n}=\frac{\sqrt{2}}{3}$.
Therefore, $(\alpha, \beta)$ can only have one solution $\left(\frac{3}{2}, \frac{\sqrt{2}}{3}\right)$.
Next, we verify that this solution satisfies the conditions.
For any $m \in \mathbf{Z}_{+}$, consider $n$ such that
$$
\frac{n(n+1)}{2}+1 \leqslant m \leqslant \frac{(n+1)(n+2)}{2} \text {. }
$$

Let $m=\frac{n(n+1)}{2}+t\left(1 \leqslant t \leqslant n+1, t \in \mathbf{Z}_{+}\right)$.
Then $\frac{\frac{1}{6} n(n+1)(n+2)}{\left[\frac{n(n+1)}{2}+n\right]^{\frac{3}{2}}} \leqslant A_{m}$
$$
\leqslant \frac{\frac{1}{6} n(n+1)(n+2)+\frac{n(n+1)}{2}}{\left[\frac{n(n+1)}{2}\right]^{\frac{3}{2}}} \text {. }
$$

Since the limits of both ends are $\frac{\sqrt{2}}{3}$, we have
$$
\lim _{n \rightarrow+\infty} A_{n}=\frac{\sqrt{2}}{3} \text {. }
$$

Hence, $(\alpha, \beta)=\left(\frac{3}{2}, \frac{\sqrt{2}}{3}\right)$.