Define the sequence $\left\{u_{n}\right\}$:
$$
u_{1}=1, u_{n+1}=\left(\sum_{k=1}^{n} u_{k}\right)^{-1} \text {. }
$$

Question: Do there exist constants $\alpha, \beta$ such that
$$
\lim _{n \rightarrow+\infty} \frac{u_{1}+u_{2}+\cdots+u_{n}}{\alpha n^{\beta}}=1 ?
$$

If they exist, find $\alpha, \beta$; if not, explain why.
Solve $\alpha=\sqrt{2}, \beta=\frac{1}{2}$.
First, prove a lemma.
Lemma $\sum_{k=1}^{n} \frac{1}{2 k-1}a_{n}^{2}+2 \text {. }$

Thus $a_{n}^{2}>2 n-1$
$\Rightarrow a_{n+1}^{2}=a_{n}^{2}+2+\frac{1}{a_{n}^{2}}<a_{n}^{2}+2+\frac{1}{2 n-1}$
$\Rightarrow a_{n}^{2}<2 n-1+\sum_{k=1}^{n-1} \frac{1}{2 k-1}(n \geqslant 2)$.
By the lemma,
$a_{n}^{2}<2 n-1+2 \sqrt{n}$.
From equations (1) and (2),
$\frac{2 n-1}{2 n}<\frac{a_{n}^{2}}{2 n}<\frac{2 n-1+2 \sqrt{n}}{2 n}$.
And $\lim _{n \rightarrow+\infty} \frac{2 n-1}{2 n}=\lim _{n \rightarrow+\infty} \frac{2 n-1+2 \sqrt{n}}{2 n}=1$, hence $\lim _{n \rightarrow+\infty} \frac{a_{n}^{2}}{2 n}=1, \lim _{n \rightarrow+\infty} \frac{a_{n}}{\sqrt{2 n}}=1$
$\Rightarrow \alpha=\sqrt{2}, \beta=\frac{1}{2}$ satisfy the conditions.
(Zhang Lei, Northeast Yucai School, Shenyang City, Liaoning Province, 110179)