In a program I am witing, a "side" calculation is related to the evaluation of the sequence $$c_n=\frac{f_{n+2}}{f_{n+1}}\qquad \text{with}\qquad n \geq 1$$ (for which an high accuracy is not required) where, with $a>1$ (this is a real number), $$f_n=\int_{0}^{\infty}\frac{dx}{\prod_{i=1}^{n}(x^a+i)}=\frac \pi{a\,(n-1)!}\,\sum _{m=0}^{n-1} (-1)^m \binom{n-1}{m}(m+1)^{\frac{1}{a}-1} \csc \left(\frac{\pi }{a}\right) $$ This does not make much problems except the high cost in terms of computer resources and numerical accuracy issues when $n$ is "large" ($n >40$) when using the formula (I can go further with numerical integration).
My work
Being short of ideas, I compared $f_n$ to $g_n$ $$g_n=\int_{0}^{\infty}\frac{dx}{(x^a+k)^n}=\frac{\Gamma \left(1+\frac{1}{a}\right) k^{\frac{1}{a}-n} \Gamma \left(n-\frac{1}{a}\right)}{\Gamma (n)}$$ and, for given $a$, I was able to numerically evaluate $k$ as a function of $n$. Over limited ranges, it became quite clear that, for a given $a$, $k\approx \alpha + \beta n$. On private, Raymond Manzoni (an MSE user) was able to show that $\color{red}{\lim_{n\to \infty } \, \beta =\frac 1e}$. The problem is, at least to me, less clear for $\lim_{n\to \infty } \, \alpha$.
Computing $$d_n=\frac{g_{n+2}}{g_{n+1}}$$ and expanding for infinitely large values of $n$, I arrived to $$d_n=\frac 1{\beta e}\left(\frac 1n-\frac 3{2n^2} \right)+O\left(\frac 1{n^3} \right)$$ and using the limit previously mentioned $$d_n=\frac 1n-\frac 3{2n^2} +O\left(\frac 1{n^3} \right)$$ This is not too bad (but not fantatstic) since $a$ does not appear until the very complicated coefficient of $n^{-3}$ while, as one could expect, it plays a non-negligible role as shown below for a few values $$\left( \begin{array}{cccc} a & n & c_n & d_n \\ 2 & 10 & 0.0819487 & 0.0850000 \\ 3 & 10 & 0.0824285 & 0.0850000 \\ 4 & 10 & 0.0826615 & 0.0850000 \\ 5 & 10 & 0.0827991 & 0.0850000 \\ 6 & 10 & 0.0828900 & 0.0850000 \\ & & & \\ 2 & 20 & 0.0451284 & 0.0462500 \\ 3 & 20 & 0.0452404 & 0.0462500 \\ 4 & 20 & 0.0452951 & 0.0462500 \\ 5 & 20 & 0.0453276 & 0.0462500 \\ 6 & 20 & 0.0453491 & 0.0462500 \\ & & & \\ 2 & 30 & 0.0311135 & 0.0316667 \\ 3 & 30 & 0.0311600 & 0.0316667 \\ 4 & 30 & 0.0311830 & 0.0316667 \\ 5 & 30 & 0.0311966 & 0.0316667 \\ 6 & 30 & 0.0312056 & 0.0316667 \\ & & & \\ 2 & 40 & 0.0236149 & 0.0240625 \\ 3 & 40 & 0.0237755 & 0.0240625 \\ 4 & 40 & 0.0237605 & 0.0240625 \\ 5 & 40 & 0.0237872 & 0.0240625 \\ 6 & 40 & 0.0237690 & 0.0240625 \end{array} \right)$$
My question
How could I do better (probably not using this tricky gymnastic) ?
Any idea and/or suggestion would be more than welcome.