How to type complex expression in Wolframalpha

Question

I tried to calculate the expression: $$\lim_{n\to\infty}\prod_{k=1}^\infty \left(1-\frac{n}{\left(\frac{n+\sqrt{n^2+4}}{2}\right)^k+\frac{n+\sqrt{n^2+4}}{2}}\right)$$ in Wolframalpha, but it does not interpret it correctly.

Could someone help me type it in and get the answer? Is it $1/2$?

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Edit: This was the AMM problem 12110, whose deadline passed on 31 August 2019.

As an alternative numerical method, I could calculate the value in MS Excel.

Answer 1

The infinite product $$ f(n) := \prod_{k=1}^\infty \left(1-\frac{n}{\left(\frac{n+\sqrt{n^2+4}}{2}\right)^k+\frac{n+\sqrt{n^2+4}}{2}}\right) \tag1$$ surprisingly can be evaluated in closed form as $$ f(n) = \frac1{2-n+\sqrt{n^2+4}}. \tag2$$ The limit of $\,f(n)\,$ as $\,n\to\infty\,$ is $\,1/2.$

If you want to use $\texttt{Mathematica}$ to find the limit, you may need to give it some help. Another answer suggests that $\, n = x - 1/x\,$ for some $\,x > 1.\,$ Using this code

 Limit[ Product[ 1 - n/(((n + Sqrt[n^2 + 4])/2)^k + (n +
       Sqrt[n^2 + 4])/2) /. (n + Sqrt[n^2 + 4]) -> 2 x /.
       n -> (x^2 - 1)/x, {k, 1, m}], x -> Infinity,
    Assumptions -> m > 1]

returns the result 1/2 in under a second. In fact, more is true. The first factor in the infinite product approaches $\,1/2\,$ as $\,x \to \infty\,$ while the other factors each approach $\,1.\,$

Answer 2

Limit[
     Product[1 - n/(((n + Sqrt[n^2 + 4])/2)^k + (n + Sqrt[n^2 + 4])/2), 
     {k, 1, \[Infinity]}],  
n -> \[Infinity]]

Answer 3

Here is the screenshot of MS Excel spreadsheet: