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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?

DinkyDoe
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1 Answers1

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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 $\lvert x\rvert \leqslant 1$.

Daniel Fischer
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