I found this on this website:
I try to generalize its conclusions:
I believe:
with $q>p>0$,
$$
\lim_{n\to+\infty}\int_0^1\cdots\int_0^1\frac{x_1^q+x_2^q+\cdots+x_n^q}{x_1^p+x_2^p+\cdots+x_n^p}dx_1dx_2\cdots dx_n=\frac{p+1}{q+1}
$$
The answers below this page give perfect proof.
We can continue to promote it.
With $f\in[0,1]$,$g\in[0,1]$,$\exists C>0 ,\forall x\in[0,1]$,$0<f(x)<Cg(x)<\infty$ and $\int_0^1g(x)dx<\infty$
$$
\lim_{n\to+\infty}\int_0^1\cdots\int_0^1\frac{\sum_{i=1}^{n}f(x_i)}{\sum_{i=1}^{n}g(x_i)}dx_1dx_2\cdots dx_n=\frac{\int_0^1f(x)dx}{\int_0^1g(x)dx}
$$
Proof of this conclusion I find it here today.
Well, this is over, and you can use this as a navigation transit page.
Best Answer
The conclusion is indeed true. Let $X_1, X_2, \ldots$ be i.i.d. random variables uniformly distributed over $[0, 1]$. Since $0 < p < q$,
$$ 0 \leq \frac{X_1^q + \cdots + X_n^q}{X_1^p + \cdots + X_n^p} \leq 1 $$
with probability one. Moreover, by the strong law of large numbers (SLLN),
$$ \lim_{n\to\infty} \frac{X_1^q + \cdots + X_n^q}{X_1^p + \cdots + X_n^p} = \frac{\lim_{n\to\infty} \frac{1}{n}(X_1^q + \cdots + X_n^q)}{\lim_{n\to\infty} \frac{1}{n}(X_1^p + \cdots + X_n^p)} = \frac{\mathbf{E}[X_1^q]}{\mathbf{E}[X_1^p]} = \frac{p+1}{q+1} $$
holds with probability one. So by the dominated convergence theorem,
$$ \int_{0}^{1} \cdots \int_{0}^{1} \frac{x_1^q + \cdots + x_n^q}{x_1^p + \cdots + x_n^p} \, \mathrm{d}x_1 \cdots \mathrm{d}x_n = \mathbf{E}\left[ \frac{X_1^q + \cdots + X_n^q}{X_1^p + \cdots + X_n^p} \right] \to \mathbf{E}\left[\frac{\mathbf{E}[X_1^q]}{\mathbf{E}[X_1^p]}\right] = \frac{p+1}{q+1}. $$
The same argument applies to the proposed generalization as well.