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$$ \sum_{n=0}^{\infty }\frac{(-1)^{n}((2n-1)!!)^2}{(2n)! (2^{2n})} = \frac{2}{\sqrt{5}} $$

I came across this summation through some other work, came across the solution as part of a function, but I was curious as to how you would solve this in a more traditional sense.

I do know it's related to the golden ratio if that helps.

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$\sum_{n=0}^{\infty }\dfrac{(-1)^{n}((2n-1)!!)^2}{(2n)! (2^{2n})} = \frac{2}{\sqrt{5}} $

Since

$\begin{array}\\ (2n-1)!! &=\prod_{k=1}^n (2k-1)\\ &=\dfrac{\prod_{k=1}^n (2k-1)\prod_{k=1}^n (2k)}{\prod_{k=1}^n (2k)}\\ &=\dfrac{(2n)!}{2^nn!}\\ \end{array} $

$\begin{array}\\ \sum_{n=0}^{\infty }\dfrac{(-1)^{n}((2n-1)!!)^2}{(2n)! (2^{2n})} &=\sum_{n=0}^{\infty }\dfrac{(-1)^{n}(\dfrac{(2n)!}{2^nn!})^2}{(2n)! (2^{2n})} \\ &=\sum_{n=0}^{\infty }\dfrac{(-1)^{n}((2n)!)^2}{(2^nn!)^2(2n)! (2^{2n})} \\ &=\sum_{n=0}^{\infty }\dfrac{(-1)^{n}(2n)!}{(n!)^2 (2^{4n})} \\ &=\sum_{n=0}^{\infty }\dfrac{(-1)^{n}}{ (2^{4n})} \binom{2n}{n}\\ \end{array} $

So look at $\sum_{n=0}^{\infty }x^n\binom{2n}{n} $ and you will get your answer.

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