How to type complex expression in Wolframalpha 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 $1/2$?

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

A: Here is the screenshot of MS Excel spreadsheet:

