Is it possible to write square root of x about c = 1 as Taylor series using sigma notation?

Question

Is it possible to write square root of x about c = 1 as Taylor series using sigma notation? I am not looking for Taylor polynomial, but Taylor series using sigma notation. I have found several questions on this forum, but no clear answer.

This is where I am right now, but sigma formula is probably not correct:

Answer 1

You can use generalized binomial for $\sqrt{x}=(1+(x-1))^\frac{1}{2}=1+\frac{1}{2}(x-1)+\frac{1}{2!}\frac{1(-1)}{2^2}(x-1)^2+\frac{1}{3!}\frac{1(-1)(-3)}{2^3}(x-1)^3+\frac{1}{4!}\frac{1(-1)(-3)(-5)}{2^4}(x-1)^4+...$

$n^{th}$ term =$\frac{1}{n!}\frac{1(-1)..(-2n+3)}{2^n}(x-1)^n$

The numerator in the $n^{th}$ term can be express as $\frac{(-1)^{n-1}(2n-2)!}{2^{n-1}(n-1)!}$ so it ends up as $\sqrt{x}=1+\sum_{n=1}^\infty \frac {1}{n!} \frac{(-1)^{n-1}(2n-2)!}{2^{2n-1}(n-1)!}(x-1)^n$.

Answer 2

To make the problem more general, let $x=y+c$ to make $$\sqrt x=\sqrt{c+y}=\sum_{n=0}^\infty \binom{\frac{1}{2}}{n} c^{\frac{1}{2}-n}y^n=\sum_{n=0}^\infty \binom{\frac{1}{2}}{n} c^{\frac{1}{2}-n}(x-c)^n$$ Truncate it to $O((x-c)^{p+1})$ if you want $p$ terms of the Taylor series.