$\lim_{N\to\infty}\frac{1}{N^aN!}\prod_{i=1}^N(a+i)$

Question

Suppose $a:=(1+\sqrt5)/2$. I want to know the value of $$\lim_{N\to\infty}\frac{1}{N^aN!}\prod_{i=1}^N(a+i).$$ I previously thought $\lim_{N\to\infty} \frac{1}{N!}\prod_{i=1}^N(a+i)$ goes to some constant value from the Stirling's approximation, but it is not true according to this link. Thus, I've been keeping looking for the asymptoic form. By taking $\log(x)$ and approximating it by integration, $$\begin{align}-a\log(N)+\int_1^N \log\left(1+\frac{a}{x}\right)dx &=a \log(a/N + 1) + \log((a/N+1)^N) + O(1)\\ &\to O(1) \text{ (as $N\to\infty$)}\\ \end{align}$$, we get constant term. Thus, I expect $\frac{1}{N!N^a}\prod_{i=1}^N(a+i)$ converges to some constant value, but I have no idea how to get that value.

Answer 1

Note that

$$\frac{1}{N^a N!}\prod_{i=1}^{N} (a+i) = \frac{1}{N^{a}\Gamma(N+1)}\frac{\Gamma(N+a+1)}{\Gamma(a+1)}.$$ We can now the following Stirling approximations: $$\log \Gamma(z) = (z-\frac12)\ln (z)-z+\frac12\ln(2\pi)+\mathcal{O}(1/z),$$ to get $$\log\frac{\Gamma(N+a+1)}{N^{a}\Gamma(N+1)}=a\log\frac{N+a+1}{N}+(N+\frac12)\log\frac{N+a+1}{N+1}-a+\mathcal{O}(1/N).$$ which is zero in the limit. Hence the required expression has $\frac1{\Gamma(a+1)}$ as a limit.

Answer 2

Gautschi's Inequality, which follows from the log-convexity of the Gamma function, says that $$ \begin{align} \lim_{n\to\infty}\frac1{n^an!}\prod_{i=1}^n(a+i) &=\lim_{n\to\infty}\frac1{n^a\Gamma(n+1)}\frac{\Gamma(n+a+1)}{\Gamma(a+1)}\\ &=\frac1{\Gamma(a+1)}\lim_{n\to\infty}\frac{\Gamma(n+a+1)}{n^a\Gamma(n+1)}\\ &=\frac1{\Gamma(a+1)} \end{align} $$