what is the easiest way   to represent $ \sqrt{1 + x} $ in series How to expand $ \sqrt{1 + x}$.
$$ \sum_{n = 0}^\infty {{\left ( 1 \over 2\right )!}x^n \over n! \left({1 \over 2 }- n\right )!} = 1 + \sum_{n = 1}^\infty {{\left ( 1 \over 2\right )!}x^n \over n! \left({1 \over 2 }- n\right )!}$$
How can I simplify $ \left({1 \over 2 }- n\right )! $?

Just to motivate a more elementary solution.
Let $$f(x)=\sqrt{x+1}=(x+1)^{1/2}$$
Now, we expand around $x=0$. 
$$f'(x)=\frac 1 2 (x+1)^{1/2-1}$$
$$f''(x)=\frac 1 2 \left(\frac 1 2-1 \right)(x+1)^{1/2-2}$$
$$f'''(x)=\frac 1 2 \left(\frac 1 2-1 \right)\left(\frac 1 2-2 \right)(x+1)^{1/2-3}$$
By induction, we get
$$f^{(n)}(x)=\prod_{k=0}^{n-1} \left(\frac 1 2 -k\right)(x+1)^{1/2-n}$$
Thus, we get
$$f^{(n)}(0)=\prod_{k=0}^{n-1} \left(\frac 1 2 -k\right)$$
$$f^{(n)}(0)=\prod_{k=0}^{n-1} \left(\frac {1 -2k}{2}\right)$$
$$f^{(n)}(0)=(-1)^{n}\frac{1}{2^{n}}\prod_{k=0}^{n-1} \left({2k-1}\right)$$
Now, note the product is exclusively of odd factors up to $(2n-3)$. What we do is fill in the missing even  numbers, and divide to keep things the same. Note that $(2n-2)\cdots 4\cdot 2=2^{n-1} (n-1)!$
$${f^{(n)}}(0) = {( - 1)^n}\frac{1}{{{2^n}{2^{n - 1}}(n - 1)!}}\prod\limits_{k = 1}^{n - 1} {\left( {2k} \right)} \prod\limits_{k = 0}^{n - 1} {\left( {2k - 1} \right)} $$
$${f^{(n)}}(0) = \frac{{{{( - 1)}^n}}}{{{2^n}{2^{n - 1}}(n - 1)!}}\prod\limits_{k = 1}^{2n - 2} k  = {( - 1)^n}\frac{1}{{{2^{2n - 1}}}}\frac{{\left( {2n - 2} \right)!}}{{\left( {n - 1} \right)!}}$$
Thus, we have that
$$\sqrt {1 + x}  = \sum\limits_{n = 0}^\infty  {\frac{{{{( - 1)}^n}}}{{{2^{2n - 1}}}}} \frac{{\left( {2n - 2} \right)!}}{{n!\left( {n - 1} \right)!}}{x^n}$$
$$\sqrt {1 + x}  = 1 + \frac{1}{2}\sum\limits_{n = 1}^\infty  {\frac{{{{( - 1)}^n}}}{{{2^{2n - 2}}}}} \frac{{\left( {2n - 2} \right)!}}{{\left( {n - 1} \right)!\left( {n - 1} \right)!}}\frac{{{x^n}}}{n}$$
$$\sqrt {1 + x}  = 1 + \frac{1}{2}\sum\limits_{n = 0}^\infty  {\frac{{{{( - 1)}^n}}}{{{2^{2n}}}}} \frac{{\left( {2n} \right)!}}{{n!n!}}\frac{{{x^{n + 1}}}}{{n + 1}}$$
$$\sqrt {1 + x}  = 1 + \sum\limits_{n = 0}^\infty  {\frac{{{{( - 1)}^{n + 1}}}}{{{2^{2n + 1}}}}} {2n\choose n}\frac{{{x^{n + 1}}}}{{n + 1}}$$
Note this gives that
$$\frac{1}{{\sqrt {1 + x} }} = \sum\limits_{n = 0}^\infty  {\frac{{{{( - 1)}^{n + 1}}}}{{{2^{2n}}}}} {2n\choose n}{x^n}$$
