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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.

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  • $\begingroup$ Expand the terms in the numerator and the denominator. What do you see? $\endgroup$ – Klangen Sep 1 '19 at 14:42
  • $\begingroup$ @Klangen I am not sure what you're trying to tell me. $\endgroup$ – Manish Kundu Sep 1 '19 at 17:17
  • $\begingroup$ The leading term above is $n^2$, while below it is $n$. And when $n\to\infty$, ... $\endgroup$ – Klangen Sep 1 '19 at 19:00
  • $\begingroup$ @Klangen so it tends to infinity for higher n, but that does not explain the oscillatory nature $\endgroup$ – Manish Kundu Sep 2 '19 at 4:39
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    $\begingroup$ As the $\ln$ is uniformly continuous on its domain then we can interverting $\lim$ and $\ln$ and the product changes to $\sum$. $\endgroup$ – Zbigniew Sep 3 '19 at 14:33
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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

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