How to find the following sum: $$\sum_{x=0}^{\infty}\left(-\frac{1}{16}\right)^x \binom{2x}{x}$$
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\begin{eqnarray*} \binom{2n}{n} = \binom{-1/2}{n}(-4)^{n}. \end{eqnarray*} \begin{eqnarray*} \sum_{n=0}^{\infty} \binom{2n}{n} \left( \frac{-1}{16} \right)^n = \sum_{n=0}^{\infty} \binom{-1/2}{n} \left( \frac{1}{4} \right)^n =\frac{1}{\sqrt{1+\frac{1}{4}}} =\color{red}{\frac{2}{\sqrt{5}}}. \end{eqnarray*}
Donald Splutterwit
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How did you derive the first and second identities? – eMathHelp Mar 04 '18 at 19:06
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For the first identity see this answer https://math.stackexchange.com/questions/2666341/how-can-i-prove-that-frac1-sqrt1-4x-sum-n-0-infty-binom-2n-n/2666404#2666404 ... the second equality of the second identity is the binomial theorem. – Donald Splutterwit Mar 04 '18 at 19:08
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@AndrewFount See this answer. – Felix Marin Mar 04 '18 at 20:21
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Due to De Moivre's identity $\cos\theta=\frac{e^{i\theta}+e^{-i\theta}}{2}$ and the orthogonality relation $\int_{0}^{2\pi}e^{ni\theta}e^{-mi\theta}\,d\theta=2\pi \delta(m,n)$ we have the following integral representation for central binomial coefficients: $$ \frac{1}{4^n}\binom{2n}{n} = \frac{2}{\pi}\int_{0}^{\pi/2}\left(\cos\theta\right)^{2n}\,d\theta\tag{1}$$ leading to $$ \sum_{n\geq 0}\frac{(-1)^n}{16^n}\binom{2n}{n}=\frac{2}{\pi}\int_{0}^{\pi/2}\sum_{n\geq 0}\frac{(-1)^n}{4^n}\left(\cos\theta\right)^{2n}\,d\theta=\frac{2}{\pi}\int_{0}^{\pi/2}\frac{d\theta}{1+\frac{1}{4}\cos^2\theta}\tag{2} $$ an by enforcing the substitution $\theta=\arctan u$ we get: $$ \sum_{n\geq 0}\frac{(-1)^n}{16^n}\binom{2n}{n} = \frac{2}{\pi}\int_{0}^{+\infty}\frac{du}{\frac{5}{4}+u^2}=\frac{2}{\sqrt{5}}.\tag{3} $$ The same approach can be used to prove the following identity: $$ \forall z\in(-1,1),\qquad \sum_{n\geq 0}\frac{z^n}{4^n}\binom{2n}{n}=\frac{1}{\sqrt{1-z}}\tag{4} $$ whose combinatorial equivalent is given by $$ \sum_{n=0}^{N}\binom{2n}{n}\binom{2N-2n}{N-n}=4^N.\tag{5} $$
Jack D'Aurizio
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