Here's one way. Start with the expansion you want, using $a_0$, $a_1$, $a_2$, $a_3$, etc. for the unknown coefficients:
$$\sqrt{x+1}\;=\;a_{0}\;+\;a_{1}x\;+\;a_{2}x^2\;+\;a_{3}x^3\;+\;a_{4}x^4\;+\;a_{5}x^5\;+ ...$$
Finding $a_0$: Plugging in $x=0$ on both sides leads to $a_{0}=1$.
Finding $a_1$: Differentiate both sides of the expansion. This gives
$$\frac{1}{2}\left(x+1\right)^{-\frac{1}{2}}\;\;=\;\;a_{1}\;+\;2a_{2}x\;+\;3a_{3}x^2\;+\;4a_{4}x^3\;+\;5a_{5}x^4\;+ ...$$
Pugging in $x=0$ on both sides leads to $a_{1}=\frac{1}{2}$.
Finding $a_2$: Differentiate 2-times both sides of the expansion. This gives
$$\left(-\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(x+1\right)^{-\frac{3}{2}}\;\;=\;\;2a_{2}\;+\;(2)(3)a_{3}x\;+\;(3)(4)a_{4}x^2\;+\;(4)(5)a_{5}x^3\;+ ...$$
Pugging in $x=0$ on both sides leads to $a_{2}=-\frac{1}{8}$.
Finding $a_3$: Differentiate 3-times both sides of the expansion. This gives
$$\left(-\frac{3}{2}\right)\left(-\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(x+1\right)^{-\frac{5}{2}}\;\;=\;\;(2)(3)a_{3}\;+\;(2)(3)(4)a_{4}x\;+\;(3)(4)(5)a_{5}x^2\;+ ...$$
Pugging in $x=0$ on both sides leads to $a_{3}=\frac{1}{16}$.
Finding $a_4$: Differentiate 4-times both sides of the expansion. This gives
$$\left(-\frac{5}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(x+1\right)^{-\frac{7}{2}}\;\;=\;\;(2)(3)(4)a_{4}\;+\;(2)(3)(4)(5)a_{5}x\;+ ...$$
Pugging in $x=0$ on both sides leads to $a_{4}=-\frac{5}{128}$.
Finding $a_5$: Differentiate 5-times both sides of the expansion. This gives
$$\left(-\frac{7}{2}\right)\left(-\frac{5}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(x+1\right)^{-\frac{9}{2}}\;\;=\;\;(2)(3)(4)(5)a_{5}\;+ ...$$
Pugging in $x=0$ on both sides leads to $a_{5}=\frac{7}{256}$.
Keep going to get as many coefficients as you want. If you keep careful track of the numbers without reducing the fractional expressions for the coefficients, you can easily determine a pattern (a pattern that can be proved by mathematical induction if you're so inclined).
What we are looking for is the Taylor series at $z=\pi/2$ for $g(z)$. The Taylor series is defined as
$$
f(z) = \sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(z-a)^n
$$
Therefore, for $g(z)$, we have
\begin{align}
g(z) &= g(\pi/2) + g'(\pi/2)(z - \pi/2) + \frac{g''(\pi/2)}{2!}(z - \pi/2)^2 + \cdots\\
&= 1 + (z - \pi/2) + \frac{1}{2!}(z - \pi/2)^2 + \frac{2}{3!}(z - \pi/2)^2 + \cdots\\
&= \sum_{n=0}^{\infty}\frac{(z - \pi/2)^n}{n!}U_{n + 1}
\end{align}
where $U_n$ are the up down numbers. You only need the first four to figure out you need up down numbers. I checked the next 3 terms as well to verify. As a note, the up down numbers are
$$
1,1,1,2,5,16,61,272,1385,\text{look up rest}
$$
In your series above, we start at $U_{n + 1}$ for $n\geq 0$.
At $z = 0$, we get $g(0) = 0$, $g'(0) = 1/2$, etc. so we yield the Taylor series for $g(z)$ at $z = 0$.
Best Answer
Comment:
In (1) we apply the binomial series expansion.
In (2) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.
In (3) we shift the index of the left-hand series by one to start with $n=1$.