How to do a singularity analysis on $M(z) := \frac{1 - z - \sqrt{1 - 2 z - 3 z^2}}{2 z}$? This question is realted to these previous questions of mine:

Use singularity analysis on
$$ M(z) := \frac{1 - z - \sqrt{1 - 2 z - 3 z^2}}{2 z}$$
to show that
$$[z^n]M(z) = 3^n \sqrt{\frac{3}{4 \pi n^3}} \biggl(1+\mathcal{O}\biggl(\frac{1}{n}\biggr) \biggr)$$

I understand now that $1/3$ is the dominant singularity of $M$. And from the lecture I know the transfer theorem and the

Standard function scale: Let $\alpha \in \mathbb{C} \setminus \mathbb{Z}_{\le 0}$. Then
$$[z^n] (1-z)^{-\alpha} = \frac{n^{\alpha-1}}{\Gamma(\alpha)} \biggl(1+\mathcal{O}\biggl(\frac{1}{n}\biggr) \biggr).$$

I can see that the formula for the standard function scale looks pretty similar to what is asked, but I do not see how to apply it here. Could you please give me a hint on how to proceed?
Edit: So far I got, using the binomial expansion of $(1-3z)^{1/2}$, that
$$(1-3z)^{1/2} = \sum_{k \ge 0} \binom{1/2}{k} 3^kz^k$$
and by the standard function scale and the fact that $\Gamma(1/2) = \sqrt{\pi}$ this entails
$$[z^n](1-3z)^{1/2} = \frac{3^n}{n^{3/2}\Gamma(1/2)} \biggl(1+\mathcal{O}\biggl(\frac{1}{n}\biggr)\biggr) = 3^n \sqrt{\frac{1}{\pi n^{3}}} \biggl(1+\mathcal{O}\biggl(\frac{1}{n}\biggr)\biggr).$$
We furthermore get
\begin{align}
[z^n]\frac{(1-3z)^{1/2}}{2z} &= \frac{3^{n-1}}{2(n-1)^{3/2}\Gamma(1/2)} \biggl(1+\mathcal{O}\biggl(\frac{1}{n-1}\biggr)\biggr) \\
&= 3^{n-1} \sqrt{\frac{1}{4\pi (n-1)^{3}}} \biggl(1+\mathcal{O}\biggl(\frac{1}{n}\biggr)\biggr).
\end{align}
 A: Comment that got too long and explains where you need to change your computations.
First $\alpha=-1/2$ in the formula so you need $|\Gamma(-1/2)|=2\sqrt \pi$ (not forgetting that $M$ has the square root with a minus, canceling the minus of $\Gamma(-1/2)$)
Second, you need to normalize the term $\sqrt {1+z}$ at $1/3$ since you have that in $M(z)$ too, and that is done by using
$$\sqrt {1+z}\sqrt {1-3z }= \biggl(\sqrt {1+z}-\frac{2}{\sqrt 3}\biggr)\sqrt{1-3z}+ \frac{2}{\sqrt 3}\sqrt {1-3z}$$
and noting that $(\sqrt {1+z}-2/\sqrt 3)=(1-3z)g(z)$ with $g$ analytic at $1/3$ so the term $(\sqrt {1+z}-2/\sqrt 3)\sqrt {1-3z}$ is of lower order and can be incorporated in the error.
Third, when you divide by $z$ you need the coeeficient of $z^{n+1}$ in the binomial expansion of $\sqrt {1-3z}$ to get the (dominant) coefficient of $z^n$ in $M(z)$ so putting it together we get that for $n \ge 1$
$$[z^n](M(z))= \frac{2}{\sqrt 3}[z^{n+1}]\frac{(1-3z)^{1/2}}{2}+ \text{error}$$ where the error above is one (negative) power of $n$ extra coming from the expansion of $(1-3z)^{3/2}$, hence it can be incorporated in the final $O(1/n)$, hence
$$[z^n](M(z)) = \frac{3^{n+1}}{2\sqrt{3 \pi} (n+1)^{3/2}}\biggl(1+O \biggl(\frac{1}{n+1}\biggr)\biggr)$$
But clearly $(n+1)^{-3/2}(1+O(1/(n+1))=n^{-3/2}(1+O(1/n))$ so one finally gets that
$$[z^n]M(z) = 3^n \sqrt{\frac{3}{4 \pi n^3}} \biggl(1+\mathcal{O}\biggl(\frac{1}{n}\biggr) \biggr)$$
