# 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.

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Do you mean to have an $n$ instead of a $k$ on the right-hand side? ($k$ is the index of the product, so it shouldn't still be around.) –  mjqxxxx Oct 11 '12 at 20:46
whoops, yes that was a typo. I fixed it. –  Haderlump Oct 11 '12 at 22:43

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 $$\prod_{k=1}^n \frac{k^\alpha}{\lambda + k^\alpha}=\mathrm e^{-s_n}= \exp\left(-\frac{\lambda}{1-\alpha}n^{1-\alpha}+o(n^{1-\alpha})\right).$$

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Thank you, did! I don't understand the estimation involving the $\max$, but one could also just calculate the integral so see that $v_n = o(n^{1-\alpha})$. Also I get different constants, eg. $u(n+1)$ and $u(0)$ instead of $u(n)$ and $u(1)$, or $w_n = \frac{\lambda}{1-\alpha}(n^{1-\alpha}-1)$ but these are just minor details that don't affect the asymptotics. –  Haderlump Oct 16 '12 at 15:50
Thanks for the appreciation. The cumbersome bound of $v_n$ is there to deal with the case $\alpha=\frac12$ but you are right that $v_n\ll n^{1-\alpha}$ is all that is needed, in the end. Both $u(n+1)$ and $u(0)$ are wrong, you might want to check again your solution (by the way, $u(0)$ does not exist). –  Did Oct 16 '12 at 15:56
Ah, yes. So how does the bound work? Do you calculate the integral in both cases and their product is then a bound? –  Haderlump Oct 17 '12 at 9:02
I do not understand your last comment, could you rephrase it? –  Did Oct 17 '12 at 9:32
I'm still trying to understand this bound $v_n=\frac12\lambda^2\int_1^nx^{-2\alpha}\,\mathrm dx\leqslant \lambda^2 C_\alpha\,\max\{n^{1-2\alpha}\log(n+1),1\}$ –  Haderlump Oct 17 '12 at 9:33

Not really a answer but a quick very informal outline:

$$\log \prod_{k=1}^n \frac{k^\alpha}{\lambda+k^\alpha } = \sum_{k=1}^n - \log(1 +\lambda \, k^{-\alpha}) \approx - \lambda \sum_{k=1}^n k^{-\alpha} \approx - \lambda \int_0^n x^{-\alpha}dx = -\lambda \frac{n^{1-\alpha}}{1-\alpha}$$

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