Asymptotic growth of $\prod_{k=1}^n \frac{k^\alpha}{\lambda + k^\alpha}$? Can you please explain why for $\lambda > 0$ and $0 < \alpha < 1$
$$\prod_{k=1}^n \frac{k^\alpha}{\lambda + k^\alpha} \sim \exp\left(-\frac{\lambda}{1-\alpha}n^{1-\alpha}+o(n^{1-\alpha})\right)$$
holds? I'm stuck at the moment.
EDIT: added $o()$ term for correctness.
 A: Define $u:x\mapsto\log(1+\lambda x^{-\alpha})$ and, for every $n\geqslant1$, $s_n=\sum\limits_{k=1}^nu(k)$. The function $u$ is decreasing hence
$$
\int_{k}^{k+1}u(x)\,\mathrm dx\leqslant u(k)\leqslant\int_{k-1}^ku(x)\,\mathrm dx.
$$
Summing these yields, for every $n$,
$$
u(n)+\int_1^nu(x)\,\mathrm dx\leqslant s_n\leqslant u(1)+\int_1^nu(x)\,\mathrm dx.
$$
For every $z\geqslant0$, $z-\frac12z^2\leqslant\log(1+z)\leqslant z$, hence
$$
w_n-v_n\leqslant\int_1^nu(x)\,\mathrm dx\leqslant w_n,
$$
where 
$$
w_n=\lambda\frac{n^{1-\alpha}}{1-\alpha},\qquad
v_n=\frac12\lambda^2\int_1^nx^{-2\alpha}\,\mathrm dx\leqslant \lambda^2 C_\alpha\,\max\{n^{1-2\alpha}\log(n+1),1\}.
$$
In particular, $v_n\ll n^{1-\alpha}$, $u(1)\ll n^{1-\alpha}$ and $u(n)\sim\lambda n^{-\alpha}\ll n^{1-\alpha}$, hence $s_n=w_n+o(w_n)$ and
$$\prod_{k=1}^n \frac{k^\alpha}{\lambda + k^\alpha}=\mathrm e^{-s_n}= \exp\left(-\frac{\lambda}{1-\alpha}n^{1-\alpha}+o(n^{1-\alpha})\right).
$$
A: Not really a answer but a quick very informal outline:
$$\log \prod_{k=1}^n \frac{k^\alpha}{\lambda+k^\alpha } = \sum_{k=1}^n - \log(1 +\lambda \, k^{-\alpha}) \approx - \lambda \sum_{k=1}^n k^{-\alpha} \approx - \lambda \int_0^n x^{-\alpha}dx = -\lambda \frac{n^{1-\alpha}}{1-\alpha}$$
