Analysis of Convergence Properties of a Series Approximation to $\sqrt{x}$ and $\frac{1}{\sqrt{x}}$

Question

I recently found a simple slowly converging infinite series approximation to $\sqrt{x}$ which may be well known, but I am struggling to prove it and fully understand its properties. The formula is $$\sqrt{x} = \sum_{k=0}^\infty \frac{1}{4^k}\binom{2k}{k}\left(1-\frac{1}{x} \right)^k$$ valid for $x>1$, with x being a rational number.

This series is zero at $x={1}$, but appears to converge to $\sqrt{x}$ for values of $x$ between $\frac{1}{2}$ and $1$.

Additional Note w.r.t. @MarkViola's Answer: With hindsight I do not need to investigate properties of above formula $x<1$ as the formula for $\frac{1}{\sqrt{x}}$ with $(x>1)$ can be found using a very similar method to MarkViola's Answer with

$$\left(1-t\right)^{1/2}=\sum_{k=0}^\infty (-1)^k\binom{1/2}{k}t^k$$

and

$$\binom{1/2}{k}=\frac{(-1)^{k-1}}{4^k \left( 2k-1\right)}\binom{2k}{k} $$

Resulting in

$$\frac{1}{\sqrt{x}} = -\sum_{k=0}^\infty \frac{1}{4^k \left( 2k-1\right)}\binom{2k}{k}\left(1-\frac{1}{x} \right)^k$$

End of Addition

The series approximation does not converge for complex numbers with real component zero, but does appear to correctly converge to $\sqrt{x}$ when calculating the square root of numbers like $(1+i)$

How do I prove the formula and in doing so best find, specify and justify the complete domain of numbers {rational, real, complex} over which such a series approximation formula to $\sqrt{x}$ will converge to the required value?

Some thoughts with Reference to @SimplyBeautifulArt's comment/question.

Introducing Catalan Number's $C_k=\frac{1}{k+1}\binom{2k}{k}$ allows the formula to be split into two parts, thus

$$\sqrt{x} = \sum_{k=0}^\infty \frac{k}{4^k}C_k\left(1-\frac{1}{x} \right)^k + \sum_{k=0}^\infty \frac{1}{4^k}C_k\left(1-\frac{1}{x} \right)^k$$

For example for $\sqrt{5}$ $$\sqrt{5} = \sum_{k=0}^\infty \frac{k}{4^k}C_k\left(1-\frac{1}{5} \right)^k + \sum_{k=0}^\infty \frac{1}{4^k}C_k\left(1-\frac{1}{5} \right)^k$$

with the ratio of the two terms seemingly giving the Golden Ratio $\phi$.

In the case of $\sqrt{2}$ the ratio of the two terms below appears to be double the Silver Ratio.

$$\sqrt{2} = \sum_{k=0}^\infty \frac{k}{4^k}C_k\left(1-\frac{1}{2} \right)^k + \sum_{k=0}^\infty \frac{1}{4^k}C_k\left(1-\frac{1}{2} \right)^k$$

Answer 1

Let $t=1-\frac1x$. Then, we have $\sqrt x =\left(1-t\right)^{-1/2}$.

Applying the generalized binomial theorem reveals

$$\left(1-t\right)^{-1/2}=\sum_{k=0}^\infty (-1)^k\binom{-1/2}{k}t^k\tag1$$

for $|t|<1$ ($x>1$).

We can write the term $(-1)^k\binom{-1/2}{k}$ as

$$\begin{align} (-1)^k\binom{-1/2}{k}&=\frac{(1/2)(3/2)(5/2)\cdots ((2k-1)/2)}{k!}\\\\ &=\frac{(2k-1)!!}{2^kk!}\\\\ &=\frac{2^k\,k!\,(2k-1)!!}{2^k\,k!\,2^k\,k!}\\\\ &=\frac{(2k)!}{4^k\,(k!)^2}\\\\ &=\frac{1}{4^k}\binom{2k}{k}\tag2 \end{align}$$

Substituting $(2)$ into $(1)$ yields the coveted result

$$\begin{align} \left(1-t\right)^{-1/2}&=\sum_{k=0}^\infty \frac{1}{4^k}\binom{2k}{k}t^k\\\\ &=\sum_{k=0}^\infty \frac{1}{4^k}\binom{2k}{k} \left(1-\frac1x\right)^k \end{align}$$

for $x>1$.