$\lim_{n\to\infty}\frac{1}{\sqrt5n!}\left(-\prod_{i=0}^{n-1}(i+\bar\phi)+\prod_{i=0}^{n-1}(i+\phi)\right)$

Question

Let $\phi:=(1+\sqrt5)/2,\bar\phi:=(1-\sqrt5)/2$ as the root of $X^2-X-1=0$. We want to know the asymptotic form of $$\begin{align} \lim_{n\to\infty}a_n=&\lim_{n\to\infty}\frac{1}{\sqrt5n!}\left(-\prod_{i=0}^{n-1}(i+\bar\phi)+\prod_{i=0}^{n-1}(i+\phi)\right). \end{align}$$ $\prod_{i=0}^{n-1}(i+\bar\phi)$is almost equal to $\Gamma(n+\bar\phi)$. So I expect we can apply the Stirling's formula, $\Gamma(z)=\sqrt{\frac{2\pi}{z}}\left(\frac{z}{e}\right)^e(1+O(1/z))$. However, I dont't know how to deal the error term arising from the fact that $\prod_{i=0}^{n-1}(i+\bar\phi)$ is not exactly equal to $\Gamma(n+\bar\phi)$. Do you have any idea to calculate $\lim_{n\to\infty} a_n$?

Answer 1

I will change the names of the variables to fit with my usual notations, so i will not type some $n$ instead of $i$, since when i would correct, typing $i$ would hurt a little.

I will write $a$ for $\phi-1$.

We fix some big $N$. Then $$ \begin{aligned} \frac 1{N!}\prod_{0\le n<N}\left(n+\phi\right) &= \frac 1{N!}\prod_{1\le n\le N}\left(n+a\right) = \prod_{1\le n\le N}\frac 1n\left(n+a\right) = \prod_{1\le n\le N}\left(1+\frac an\right) \\ &\ge 1+\sum_{1\le n\le N}\frac an \end{aligned} $$ and the last expression goes to infinity. Doing the same with the other product, we would introduce $b= \bar\phi-1$ which is negative, so most (all but some first) factors are between zero and one, so the second product goes to zero.

The limit in the OP is thus infinity.