= Solution
Use the <generalized binomial theorem> with exponent $-1/2$, choosing the <analytic branch of a square root> that equals $1$ at $z=0$. For $|z|<1$,
$$
(1-z)^{-1/2}=\sum_{n=0}^{\infty}\binom{-1/2}{n}(-z)^n.
$$
The signs cancel in each <binomial coefficient>. For $n\geq1$,
$$
(-1)^n\binom{-1/2}{n}=\frac{1\cdot3\cdots(2n-1)}{2^n n!}=\frac{(2n)!}{2^{2n}(n!)^2}=\frac1{4^n}\binom{2n}{n}.
$$
The same expression equals $1$ for $n=0$, using the empty-product convention. Consequently the <power series> is
$$
\boxed{\frac1{\sqrt{1-z}}=\sum_{n=0}^{\infty}\binom{2n}{n}\left(\frac z4\right)^n,\qquad |z|<1.}
$$
Its coefficients are the <normalized central binomial coefficients>. As a convergence check, the ratio of successive absolute terms tends to $|z|$, so the <ratio test> gives absolute convergence inside the <unit disc>. Applying the <Cauchy coefficient formula> to the <holomorphic function> $(1-z)^{-1/2}$ now immediately yields
$$
\boxed{\int_{\gamma(0,r)}\frac{dz}{z^{n+1}\sqrt{1-z}}=\frac{2\pi i}{4^n}\binom{2n}{n},\qquad 0<r<1,\quad n=0,1,2,\ldots.}
$$
This value is independent of the chosen <radius> because the <contour integral> extracts the same coefficient on every such <circle>.
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