<s>[INST] How to expand this taylor series and find radius of convergence f(x)= √(1-x) at x=0
How do you find the taylor series and radius of convergence?
 [/INST] $$(1-x)^\frac{1}{2}=1+\sum_{k=1}^\infty(-1)^k\frac{\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)\cdots(\frac{1}{2}-k+1)}{k!}x^k$$
and
\begin{align}\frac{\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)\cdots(\frac{1}{2}-k+1)}{k!}&=\frac{(-1)(-3)\cdots(3-2k)}{2^kk!}=(-1)^{k-1}\frac{1\times3\times\cdots\times(2k-3)}{2^kk!}\\&=(-1)^{k-1}\frac{(2k-2)!}{2^{2k-1}k!(k-1)!}\end{align}
hence we have
$$(1-x)^\frac{1}{2}=1-\sum_{k=1}^\infty \frac{(2k-2)!}{2^{2k-1}k!(k-1)!}x^k=1-\sum_{k=1}^\infty a_kx^k$$
Finaly by the D'Alembert criterion we have
$$\frac{1}{R}=\lim_{k\to\infty}\left|\frac{a_{k+1}}{a_k}\right|=\lim_{k\to\infty}\frac{2k(2k+1)}{2^2(k+1)k}=1$$
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