Considering $$A(z)=\log(Y(z))=\log \left(\Gamma \left(\frac{1}{2}+z\right)\right)-\log (\Gamma (z+1))$$ you are looking for the Taylor expansion built around $z=0$. This means $$A(z)= A(0)+\frac{A'(0)} {1!}z+\frac{A''(0)} {2!}z^2+\frac{A'''(0)} {3!}z^3+\cdots$$
You are given for the derivatives
$$A'(z)=\psi ^{(0)}\left(z+\frac{1}{2}\right)-\psi ^{(0)}(z+1)$$
which generalizes to
$$A''(z)=\psi ^{(1)}\left(z+\frac{1}{2}\right)-\psi ^{(1)}(z+1)$$
$$A'''(z)=\psi ^{(2)}\left(z+\frac{1}{2}\right)-\psi ^{(2)}(z+1)$$
which, now, need to be simplified using $z=0$.
So $$A(0)=\frac{\log (\pi )}{2}\qquad A'(0)=-\log (4)\qquad A''(0)=\frac{\pi ^2}{3}\qquad A'''(0)=-12 \zeta (3)$$ This makes $$A(z)=\frac{\log (\pi )}{2}-\log (4)z+\frac{\pi ^2 }{6}z^2-2 \zeta
(3)z^3+O\left(z^4\right)$$ Now, Taylor again $$Y(z)=e^{A(z)}=\sqrt{\pi }-\sqrt{\pi } \log (4)z+\frac{1}{6} \sqrt{\pi } \left(\pi ^2+3 \log
^2(4)\right)z^2-\frac{1}{6} \left(\sqrt{\pi } \left(12 \zeta (3)+\log ^3(4)+\pi
^2 \log (4)\right)\right)z^3 +O\left(z^4\right)$$ Now, to finish, replace $z$ by $\frac 1 {2n}$.
Just to see how accurate is the expansion,
$$\left(
\begin{array}{ccc}
n & \text{exact} & \text{approximation}\\
5 & 1.565345082 & 1.563836840 \\
10 & 1.660110105 & 1.660007284 \\
15 & 1.695365291 & 1.695344347 \\
20 & 1.713776688 & 1.713769957
\end{array}
\right)$$