Convergence of Taylor series of $\sqrt{1-x}$ Concerning $$\sqrt{1-x} = \sum_{k=0}^{\infty} \left[\prod_{j=1}^k \left(\frac{j-1-\frac{1}{2}}{j}\right)\right]x^k$$ 
the Taylor series about $x=0$. For $|x|< 1$ this series converges uniformly. 
It appears, I read here http://www.math.purdue.edu/~zhang24/StoneWeierstrass.pdf
that the series converges uniformly for $x\in [0,1]$
So how can we show this? We need to know then also that it converges for $x=1$, and then we need to see how these coefficients of $x^k$ must be summable, i guess..
I just don't see how?

The terms
$$a_k = (-1)^k\binom{\frac{1}{2}}{k} = \prod_{j=1}^k \frac{j-1-\frac{1}{2}}{j}$$
satisfy
$$\left\lvert \frac{a_{k+1}}{a_k}\right\rvert = \frac{(k+1)-1-\frac{1}{2}}{k+1} = 1 - \frac{\frac{3}{2}}{k+1}.$$
By Raabe's test, the sum is absolutely convergent for $\lvert x\rvert = 1$.
The uniform convergence for $\lvert x\rvert \leqslant 1$ then follows, since
$$\left\lvert \sum_{k=N}^\infty (-1)^k\binom{\frac{1}{2}}{k}x^k\right\rvert \leqslant \sum_{k=N}^\infty \left\lvert \binom{\frac{1}{2}}{k}\right\rvert$$
for all $\lvert x\rvert \leqslant 1$.
