Sum to infinity

Relevant wiki .- fractional binomial theorem

The power series ("McLaurin series" in $\mathbb{C}$ ) or analytic expansion of $(1 + 2x)^{\frac{-1}{2}} = \frac{1}{\sqrt{1 + 2x}}$ with $|x| < 1/2$ is : $\displaystyle \sum_{k = 0}^{\infty} {-1/2 \choose k} (2x)^k = 1 + (-1/2)\cdot(2x) + \frac{(-1/2)\cdot (-3/2)}{2!} (2x)^2 + \frac{(-1/2)\cdot(-3/2)\cdot(-5/2)}{3!}(2x)^3 +... =$ Now substituing $x = 1/8$ we get $1 + (-1/2)\cdot (1/4) + \frac{1\cdot 3}{2! \cdot 4 \cdot 16} - \frac{1\cdot 3\cdot 5}{3! \cdot 8 \cdot 64} + ... =$ $= 1 - 1/8 + \frac{1\cdot 3}{ 8 \cdot 16} - \frac{1\cdot 3\cdot 5}{8 \cdot 16 \cdot 24} + ... =$ $= (1 + 2\cdot \frac{1}{8})^{\frac{-1}{2}} = \frac{1}{\sqrt{1 + \frac{1}{4}}} = \frac{2}{\sqrt{5}}$

Details.-

$8 = 8^1 \cdot 1!, \quad \\ 8 \cdot 16 = 8^2 \cdot 2!, \quad \\8 \cdot 16 \cdot 24 = 8^3 \cdot 3!...$

if $\alpha \in \mathbb{C}$ and $n \in \mathbb{Z}^+ = \{1,2,3,...\}$ then ${\alpha \choose n} = \frac{\alpha \cdot (\alpha - 1) \cdot \cdot \cdot (\alpha - n + 1)}{n!}$ and ${\alpha \choose 0} = 1$