Sum up the series!

Relevant wiki .- fractional binomial theorem

The power series of $(1 - 2x)^{\frac{-1}{2}} = \frac{1}{\sqrt{1 - 2x}}$ with $|x| < 1/2$ is : $1 + x + \frac{3x^2}{2} + \frac{5x^3}{2} + \frac{35x^4}{8} + \frac{63x^5}{8} + o(x^5) = \displaystyle \sum_{k = 0}^{\infty} {-1/2 \choose k} (-2x)^k =$ $=1 + (-1/2)\cdot(-2x) + \frac{(-1/2)\cdot (-3/2)}{2!} (-2x)^2 + \frac{(-1/2)\cdot(-3/2)\cdot(-5/2)}{3!}(-2x)^3 +....$ Now substituing $x = 1/3$ we get $1 + \frac{1}{3} + \frac{1\cdot 3}{3\cdot 6} + \frac{1\cdot 3\cdot 5}{3\cdot 6\cdot 9} + ... =$ $(1 - 2\cdot \frac{1}{3})^{\frac{-1}{2}} = \frac{1}{\sqrt{1 - 2\cdot\frac{1}{3}}} = \sqrt{3}$ $Y = 3 \Rightarrow 5Y + 1 = 16$

Details.-

$3 = 3^1 \cdot 1!, \quad \\ 3 \cdot 6 = 3^2 \cdot 2!, \quad \\3 \cdot 6 \cdot 9 = 3^3 \cdot 3!...$