sigma notation of Taylor series of $\frac{1}{\sqrt{x}}$

Question

I am having trouble finding the sigma notation for the taylor series of $f(x) = \frac{1}{\sqrt{x}}$ with the bound $b = 1$. I believe I found a solution but I would like to find one that doesn't involve a double factorial.

This is the solution I found: $$T_n(x) =\sum_{k=0}^{n} (-1)^k\frac{(2k-1)!!}{k!\cdot2^k}(x-1)^k$$

Answer 1

Probably the simplest notation could be $$T_n(x)=\sum_{k=0}^n \binom{-\frac{1}{2}}{k}(x-1)^k$$

Answer 2

Given the well-known $$\frac{1}{\sqrt{1-4x}}=\sum_{k=0}^\infty \binom{2k}{k} x^k,$$ note that $$\frac{1}{\sqrt{x}}=\frac{1}{\sqrt{1-4(1-x)/4}}$$ and perform the change of variable $x\mapsto (1-x)/4$ to obtain $$\frac{1}{\sqrt{x}}=\sum_{k=0}^\infty \binom{2k}{k} \left(\frac{1-x}{4}\right)^k=\sum_{k=0}^\infty \frac{\binom{2k}{k}}{(-4)^k} (x-1)^k.$$

Answer 3

$$\frac{(2k-1)!!}{2^k}=\frac{\Gamma\left(k-\frac12\right)}{\sqrt\pi}.$$

You can also absorb the sign with

$$(-1)^k\frac{(2k-1)!!}{2^k}=\frac{\sqrt\pi}{\Gamma(\frac12-k)}.$$