( $\frac12$ )/2+( $\frac12$ )( $\frac34$ )/3+( $\frac12$ )( $\frac34$ )( $\frac56$ )/4+...= ?

$\int_0^1{\left(1 - x\right)}^{-\frac12} d x$ = [ $\frac{C_0}{1} x - \frac{C_1}{2} x^2 + \frac{C_2}{3} x^3 - \frac{C_3}{4} x^4 + \ldots$ ] $_0^1$

$\Rightarrow 2 = 1$ + $\displaystyle \frac{\frac12}{2}$ + $\displaystyle \frac{\frac12 \frac34}{3}$ + $\displaystyle \frac{\frac12 \frac34 \frac56}{4}$ + $\displaystyle \frac{\frac12 \frac34 \frac56 \frac78}{5}$ + $\displaystyle \ldots$

$\Rightarrow$ $\boxed 1$ = $\displaystyle \frac{\frac12}{2}$ + $\displaystyle \frac{\frac12 \frac34}{3}$ + $\displaystyle \frac{\frac12 \frac34 \frac56}{4}$ + $\displaystyle \frac{\frac12 \frac34 \frac56 \frac78}{5}$ + $\displaystyle \ldots$

One piece of Cayley's sum is proven up to here. Together with further extra completion:

$\int_0^1 \frac{{\left (1 - x\right)}^{-\frac12} - C_0}{x} d x$ = [ $- \frac{C_1}{1} x + \frac{C_2}{2} x^2 - \frac{C_3}{3} x^3 + \frac{C_4}{4} x^4 - \ldots$ ] $_0^1$

$\int_0^1 \frac{{\left (1 - x\right)}^{-\frac12} - 1}{x} d x$ = $\displaystyle \frac{\frac12}{1}$ + $\displaystyle \frac{\frac12 \frac34}{2}$ + $\displaystyle \frac{\frac12 \frac34 \frac56}{3}$ + $\displaystyle \frac{\frac12 \frac34 \frac56 \frac78}{4}$ + $\displaystyle \ldots$

$\int_{\epsilon}^{1 - \epsilon} \frac{{\left (1 - x\right)}^{-\frac12} - 1}{x} d x$ = 1.386294361119+ = Ln 4 with $\epsilon$ = 7.013825370499E-08

using 1,000,000 strips of parabolic curves. You may evaluate in non-numerical method.

The two sums of Cayley's series are proven with this definite integral completed. Not less than a million terms are required with a 15 significant figure calculator! Note that numerical integral is much faster than series summation with $\epsilon$ to play a trick!

When detached, $B$ ( $\frac12$ , 0) - (- $-\infty$ ) = $\Gamma$ (0) - $\infty$ = $\infty$ - $\infty$ = {indeterminate}

With non-numerical approach, let $x = \sin^2 \theta$ :

Final integral = 2 $[Ln$ $\displaystyle \frac {1}{2 cos^2 \frac {\theta }{2}}$ $\displaystyle ]_0^{\frac {\pi}{2}}$ = 2 [Ln 1 - Ln $\frac12$ ] = $Ln$ $4$

$\displaystyle \boxed {Proof~Of~Cayley's~Sums~Completed!}$