Furthermore, *log*-convexity closes under multiplication. This implies that $G_m^n := \Gamma(m+\beta)\Gamma(n-m+\beta)$ is a log-convex function on the interval $m \in [1,n-1]$. A convex function always attains its maximum on the boundary of its domain. Accordingly, $G_m^n \leq G_{1}^n$ for $m \in [1,n-1]$. This way, the sum $\sum_{m=0}^{n}G_m^n$ can be estimated after stripping off the two boundary terms: $$\begin{aligned}
 \label{eqn:sumGestimate} &\sum_{m=0}^{n} G_m^n \leq 2 G_{0}^n + (n-1) G_{1}^n \leq 2 G_{0}^n + (n-1+\beta) G_{1}^n & & \forall n\geq1. \end{aligned}$$ It follows from $n \Gamma(n) = \Gamma(n+1)$ that $ G_{1}^n = G_{0}^n \frac{\beta}{n-1+\beta} $ for all $n\geq1$. Substituting this into eq.  gives the estimate in eq.  with $C=(2+\beta)\Gamma(\beta)$. The remaining case $n=0$ is trivial.

\[crll:centersum\] If $\beta \in {\mathbb{R}}_{>0}$, then $$\begin{aligned}
 \sum_{m=R}^{n-R} \Gamma(m+\beta) \Gamma(n-m+\beta) & \in {\mathcal{O}}(\Gamma(n-R+\beta)) && \forall R \in {\mathbb{N}}_0. \end{aligned}$$

Rewrite the left hand side as $\sum_{m=0}^{n-2R} \Gamma(m+R+\beta) \Gamma(n-R-m+\beta)$. Lemma \[lmm:gammasum\] can be applied with the substitutions $\beta \rightarrow \beta+R$, $n\rightarrow n-2R$ to obtain the required estimate.

\[crll:exponential\_inO\] If $\beta \in {\mathbb{R}}_{>0}$, $C \in {\mathbb{R}}$ and $P \in {\mathbb{R}}[m]$ some polynomial in $m$, then $$\begin{aligned}
 \sum_{m=R}^{n} C^m P(m) \Gamma(n-m+\beta) \in {\mathcal{O}}(\Gamma(n-R+\beta)) && \forall R \in {\mathbb{N}}_0. \end{aligned}$$

There is a $C' \in {\mathbb{R}}$ such that $|C^m P(m)|$ is bounded 