Analysis of Convergence Properties of a Series Approximation to $\sqrt{x}$ and $\frac{1}{\sqrt{x}}$ 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.
$$\sqrt{2} = \sum_{k=0}^\infty \frac{k}{4^k}C_k\left(1-\frac{1}{2} \right)^k + \sum_{k=0}^\infty \frac{1}{4^k}C_k\left(1-\frac{1}{2} \right)^k$$

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 $x>1$.
