Evaluating $\lim_{n \to \infty} \prod_{i=0}^{n-1} \left(\frac{n(n+1)}{2(n-1)} - i\right)^{n-i}$ I am trying to find this limit. 
$$\lim_{n \to \infty} \prod_{i=0}^{n-1} \left(\frac{n(n+1)}{2(n-1)} - i\right)^{n-i}$$
I am not sure how to proceed. I tried to check out the values of the product for increasing values of $n$ and it looks like the limit oscillates between $+\infty$ and $-\infty$ for every 4 terms, so I believe the limit does not exist. I am not sure how I can prove that. 
 A: We define $g,h:\mathbb{N} \to \mathbb{N} $ with 
$g(x)=\#\{i \in \{0,x-1\}:\frac{n(n+1)}{2(n-1)}-i>0 \}$
and $h(x)=\#\{i \in \{0,x-1\}:\frac{n(n+1)}{2(n-1)}-i<0 \}$
$\forall n>3 :\frac{n(n+1)}{2(n-1)} \notin \mathbb{Z}$
and $\exists n_0>3 \in \mathbb{N} \forall n \geq n_0 :n/2+1<\frac{n(n+1)}{2(n-1)}<n/2+1+1/3$
As a result :
$\forall n\geq n_0 : n=g(n)+h(n)$ and by a  combinatorial argument
$g(4n)=2n+2 , h(4n)=2n-2 , g(4n+2)=2n+3 , h(4n+2)=2n-1$
We now have that :
$\forall n \geq n_0 : \prod _{i=0}^{4n}\left ( \frac{4n(4n+1)}{2(4n-1)}-i \right )^{4n-i}>0 $ and $\prod _{i=0}^{(4n+2)+2}\left ( \frac{(4n+2)((4n+2)+1)}{2((4n+2)-1)}-i \right )^{(4n+2)-i}<0 $
Thus the oscillatory nature.
Also :
$min\{\left| \frac{2n(2n+1)}{2(2n-1)}-i \right|^{2n-i}| : i \in \{0,2n-1\}\} = \left|\frac{2n(2n+1)}{2(2n-1)}-(n+1) \right |^{2n-(n+1)} = |\frac{2}{2n-1}|^{n-1} $
So 
$\prod _{i=0}^{2n}\left | \frac{2n(2n+1)}{2(2n-1)}-i \right |^{2n-i} \geq 
|\frac{2}{2n-1} |^{n-1} |\frac{n(2n+1)}{2n-1}|^{2n} \to +\infty$
And your statement is proved
