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I am trying to understand/prove a lemma which is stated in two papers about branching processes (Lemma 2.2 in "Martingales And Large Deviations For Binary Search Trees" by Jabbour-Hattab and Lemma 2.8 in "Martingales, Embedding and Tilting of Binary Trees" by Chauvin). The papers refer to the Stirling Formula and standard analytic methods without giving more insight in the proof.

We are given a product $C_n(z)=\prod_{k=0}^{n-1}\frac{k+2z}{k+1}$ for $n\geq 1$ and $C_0(z)=1$, and the lemma states that for positive $z$, we have the following asymptotic behaviour when $n$ tends to infinity:

$C_n(z) \sim \frac{n^{2z-1}}{\Gamma(2z)}$

Unfortunately, I don't have a lot of knowledge when it comes to the Stirling formula or the Gamma function. Any help would be much appreciated.

CampFire
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  • $C_n\sim\frac{n^{2z-1}}{\Gamma(2z+1)}$ (you can check, for example, for $z=0$) $$C_n(z)=\prod_{k=1}^{n-1}\frac{k+2z}{k+1}=\frac{(2 z+1)...(2z+n-1)}{n!},\frac{\Gamma(2z+1)}{\Gamma(2z+1)}=\frac{\Gamma(n+2z)}{\Gamma(n+1)\Gamma(2z+1)}$$ $$=\frac{e^{\ln\Gamma(n+1+2z-1)}}{\Gamma(n+1)\Gamma(2z+1)}=\frac{e^{\ln\Gamma(n+1)+\Psi(n+1)\cdot(2z-1)+\Psi^{(1)}(n+1)\frac{(2z-1)^2}{2!}+...}}{\Gamma(n+1)\Gamma(2z+1)}$$ Given that $\Psi(n+1)=\ln n+O(1/n)$ and $\Psi^{(1)}(n+1)\sim O(1/n)$ $$C_n=\frac{e^{\ln\Gamma(n+1)+(2z-1)\ln n+O(1/n)}}{\Gamma(n+1)\Gamma(2z+1)}=\frac{n^{2z-1}e^{O(1/n)}}{\Gamma(2z+1)}$$ – Svyatoslav May 11 '24 at 17:15

1 Answers1

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We notice that $\prod_{k=0}^{n-1}\frac{k+2z}{k+1}=\frac{2z\cdot(2z+1)\cdot \dots \cdot (2z+n_-1)}{1\cdot 2\cdot \dots \cdot (n-1)\cdot n}=\frac{1}{n!}\cdot \prod_{k=0}^{n-1}(k+2z).$

Now, we know that for $n\in \mathbb{N}$, $\Gamma(n+1)=n!.$

Moreover, we have that $\frac{\Gamma(n+x)}{\Gamma(x)}=x\cdot(x+1)\cdot \dots \cdot (x+n-1).$

So, we get that $C_n(z)= \frac{1}{\Gamma(n+1)}\cdot \frac{\Gamma(2z+n)}{\Gamma(2z)}=\frac{\Gamma(2z-1+n+1)}{\Gamma(2z)\cdot \Gamma(n+1)}.$

Now, using that $\Gamma(x+\alpha)\sim \Gamma(x)\cdot x^{\alpha}$ for $x\to \infty$ and $\alpha \in \mathbb{C}$ (the proof for this can be found here), we get that

$C_n(z)=\frac{\Gamma(2z-1+n+1)}{\Gamma(2z)\cdot \Gamma(n+1)} \sim \frac{n^{2z-1}}{\Gamma(2z)}\sim \frac{(n+1)^{2z-1}}{\Gamma(2z)}.$

CampFire
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