Thank you. To me it also looks canonical - one could e.g. go easily go on above and show the local limit theorem asymptotics. I saw the simpler derivation (your suggestion) only in retrospect, but have added an explanation now.

Meanwhile the paper has been digitized by google: books.google.de/books/about/…

For $p=1$ and general $q$ formulas similar to the case $p=1,q=2$ exist, as in the linked answer one can show that $$\mathbb{E}\bigg(\frac{X_1^q+\ldots + X_{n+1}^q}{X_1+\ldots+X_{n+1}}\bigg)=\frac{n(n+1)}{n!} \sum_{i=0}^n {n \choose i} (-1)^{n-i} \int_0^1 u^q(u+i)^{n-1}\log(u+i)\,du$$, and one proceeds from here simarly as there.

@mathworker21: not much thought, really. I thought about constructing (highly dependent) uniform marginals using Dirichlet distributions (and was lucky at the first try).

@mathworker21: the following example shows that asymptotically the "truth" is near the lower bound given by Jensen. Let $(Y_1,\ldots,Y_k)$ be uniform on the $k-1$-dimensional simplex $\mathcal{S}_{k-1}:=\{(x_1,\ldots,x_k)\in [0,1]^k\,\mid \sum_{i=1}^k x_i=1\}$, and let $X_i:=(1-Y_i)^{k-1}$. Then each $X_i$ is uniform on $[0,1]$ and $$d_k:=\mathbb{E}\prod_{i=1}^k X_i=\mathbb{E}\prod_{i=1}^k (1-Y_i)^{k-1}\leq \big((1-\frac{1}{k})^k\big)^{k-1}\approx e^{\tfrac{1}{2}-k} $$ since by the AGM-inequality $\prod_{i=1}^k (1-Y_i)\leq (1-\frac{1}{k})^k$.

The setup is from the "Pill problem" (AMM-problem E3429,(vol. 98(3) 1991, p.264) by Knuth and McCarthy)): A certain pill bottle contains m large pills and n small pills, where each large pill is equivalent to two small ones. Each day the patient chooses a pill at random, if a small pill is selected, (s)he eats it; otherwise (s)he breaks the selected pill and eats one half, replacing the other half, which is thenceforth considered to be a small pill. --You'll find some literature using this keyword.

Luckily, the upper bound has a rapid proof. $$I(n):=\int_0^\infty (1+\frac{x}{n})^n\,e^{-x}\,dx=n\,\int_{0}^{\infty}(1+y)^n e^{-ny}\,dy=n\int_0^\infty e^{-n\,\frac{z^2}{2}}\frac{z}{y(z)}(1+y(z))\,dz$$ where in the last step the substitution $y=y(z)$ was used, with $z(y):=\sqrt{2\,\left(y-\log(1+y)\right)}$ (and y(z) it's inverse). Now $\frac{z(y)}{y}\leq 1$, hence $I(n)\leq n\int_0^\infty e^{-n\,\frac{z^2}{2}}(1+z)\,dz=\sqrt{\frac{\pi}{2}\,n}+1\;.$

Upper bound for the second integral: in AMM problem 11353 (solution in AMM 117 (January 2010)) it is shown that $$\frac{2}{3}<\int_{0}^\infty(1+\frac{x}{n})^n e^{-x}\,dx -\sqrt{\frac{\pi}{2}n}<1$$