Bounding Error term in Taylor Expansion of $\sqrt{1+x}$ I am attempting to justify the expansion
$$
\sqrt{1+x}= 1 + \frac{x}{2} + \sum_{n=2}^{\infty}{(-1)^n \frac{1}{2n}\frac{(1-\frac{1}{2}) \cdots ((n-1)-\frac{1}{2})}{(n-1)!}x^n}
$$
for $-1&ltx\leq 1$
I've got the expansion, but I cannot prove that the error term tends to zero. 
$$
E_n = \frac{(-1)^{n-1}(2n-3)(2n-1) \cdots (1)}{2^n n!}(1+\theta x)^{-\frac{2n-1}{2}}x^n
$$
where $\theta \in (0,1)$
The question suggests using the Constancy Lemma (if the differential is zero, the function is constant), but I can't make that work either. Any help gratefully appreciated.
(While technically this is homework I'm the tutor so I allowed to cheat! Also it is extra embarrassing that I cannot do it)
 A: Consider the absolute value 
$$
a_n=\frac{1}{2n}\frac{(1-\frac{1}{2}) \cdots ((n-1)-\frac{1}{2})}{(n-1)!}
$$ of the coefficient of $x^n$ in  the series expansion of $\sqrt{1+x}$. Then 
$$
\frac{a_{n+1}}{a_n}=\frac{n-\frac12}{n+1}=1\color{red}{-\frac32}\frac1n+o\left(\frac1n\right),
$$ 
and the heuristics in such cases is that $a_n$ behaves roughly like $n^s$ with $s=\color{red}{-\frac32}$ since, for any $s$, 
$$
\frac{(n+1)^s}{n^s}=1+\frac{s}n+o\left(\frac1n\right).
$$
To continue the proof more rigorously, choose any $t\lt\frac32$ and consider $b_n=n^ta_n$. Then 
$$
\frac{b_{n+1}}{b_n}=\frac{(n+1)^t}{n^t}\frac{a_{n+1}}{a_n}=\left(1+\frac{t}n+o\left(\frac1n\right)\right)\cdot\left(1-\frac32\frac1n+o\left(\frac1n\right)\right),
$$
hence
$$
\frac{b_{n+1}}{b_n}=1+\left(t-\frac32\right)\frac1n+o\left(\frac1n\right).
$$
Since $t-\frac32\lt0$, this implies that $(b_n)$ is ultimately decreasing, in particular $b_n\leqslant c_t$ for every $n$, for some finite $c_t$.
Finally, $a_n\leqslant c_tn^{-t}$ uniformly over $n$ for every $t\lt\frac32$,
and one can choose $t\gt1$. Then $\sum\limits_n n^{-t}$ converges (absolutely) hence $\sum\limits_na_n$ converges (absolutely) and the series expansion of $\sqrt{1+x}$ converges (absolutely) for every $|x|\leqslant1$.
