= Solution
The <Taylor theorem with Lagrange remainder> says that, if $f$ has $N+1$ <continuous> derivatives on the interval joining $0$ and $x$, then for some $\xi$ strictly between them,
$$
f(x)=\sum_{j=0}^N\frac{f^{(j)}(0)}{j!}x^j+
\frac{f^{(N+1)}(\xi)}{(N+1)!}x^{N+1}.
$$
More generally the expansion is about any base point, replacing $x$ by its displacement. The displayed regularity is a sufficient hypothesis for the theorem.
For $f(t)=(1+t)^{1/2}$ and $t>-1$, repeated differentiation gives
$$
f^{(n)}(t)=\left[\frac12\left(\frac12-1\right)\cdots\left(\frac12-n+1\right)\right](1+t)^{1/2-n}.
$$
Therefore the requested coefficients are $c_n=f^{(n)}(0)/n!$. For a fixed $0<x<1$, the remainder has
$$
|R_N(x)|=|c_{N+1}|(1+\xi)^{-N-1/2}x^{N+1}\leq |c_{N+1}|x^{N+1},\qquad0<\xi<x.
$$
Now $|c_1|=1/2$ and $|c_{n+1}/c_n|=(n-1/2)/(n+1)<1$ for $n\geq1$, so $|c_{N+1}|\leq1/2$. Hence $|R_N(x)|\leq x^{N+1}/2\to0$. Passing to the limit proves, rather than merely formally suggests, the <binomial series>
$$
\boxed{\sqrt{1+x}=1+\sum_{n=1}^\infty c_nx^n\qquad(0<x<1).}
$$
The restriction $x>0$ makes the remainder bound especially direct, since $1+\xi\geq1$.
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