= Solution
For the first <derivative>, $f'(x)=\tfrac12(1+x)^{-1/2}$, which is the stated formula at $r=1$. If the formula holds at $r$, differentiating multiplies its coefficient by $\tfrac12-r=-(2r-1)/2$. The factorial coefficients satisfy
$$
\frac{(2r-2)!}{2^{2r-1}(r-1)!}\frac{2r-1}{2}
=\frac{(2r)!}{2^{2r+1}r!}.
$$
The sign changes, proving by <mathematical induction> that
$$
\boxed{f^{(r)}(x)=(-1)^{r-1}\frac{(2r-2)!}{2^{2r-1}(r-1)!}
(1+x)^{1/2-r}\qquad(r\geq1).}
$$
Apply the given <Taylor formula with integral remainder> at $x=1/2$. Its polynomial part is $1+2S_n$, where $S_n$ is the first $n$ terms of the requested series, since
$$
\frac{f^{(r)}(0)}{r!}2^{-r}
=2(-1)^{r-1}\frac{(2r-2)!}{8^r r!(r-1)!}.
$$
To justify passage to the infinite <series>, for $0\leq t\leq1/2$,
$$
|f^{(n+1)}(t)|\leq\frac{(2n)!}{2^{2n+1}n!}.
$$
Therefore the integral remainder obeys
$$
\begin{aligned}
|R_n(1/2)|
&\leq\frac{(2n)!}{2^{2n+1}(n!)^2}
\frac{2^{-(n+1)}}{n+1}\\
&=\frac{\binom{2n}{n}}{2^{3n+2}(n+1)}
\leq\frac{2^{-n-2}}{n+1}\longrightarrow0.
\end{aligned}
$$
The last inequality follows from $\binom{2n}{n}\leq\sum_{j=0}^{2n}\binom{2n}{j}=4^n$, a direct <binomial theorem> bound. Hence $\sqrt{3/2}=1+2\lim S_n$, and
$$
\boxed{\sum_{r=1}^{\infty}(-1)^{r-1}\frac{(2r-2)!}{8^r r!(r-1)!}
=\frac{\sqrt6-2}{4}.}
$$
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