Using the binomial theorem

Question

I'd like some help with proving the next equation:

$$\sqrt{1+x}=\sum_{0}^{\infty }\frac{(-1)^{n-1}}{2^{2n-1}\cdot n}\binom{2n-2}{n-1}\cdot x^{n}$$

Answer 1

For $|x|<1$ we have (Newton)$$(1+x)^{1/2}=1+x(1/2)/1!+x^2(1/2)(1/2-1)/2!+x^3(1/2)(1/2-1)(1/2-2)/3!+...$$ The coefficients $A_n$ of $x^n$ in this series satisfy the conditions $A_0=1$ and $A_{n+1}=A_n(1/2-n)/(n+1).$

Defining $\binom {-2}{-1}=1$ (so the term with $n=0$ in your series makes sense) you can confirm that the coefficients in your series satisfy these conditions also so it is exactly Newton's series.

Answer 2

Let's rewrite the expression in terms of the Gamma function $$ \sqrt {1 + x} = \sum\limits_{0 \leqslant n} {\frac{{\left( { - 1} \right)^{\,n - 1} }} {{2^{\,2n - 1} n}}\left( \begin{gathered} 2(n - 1) \\ n - 1 \\ \end{gathered} \right)x^{\,n} } = \sum\limits_{0 \leqslant n} {\frac{{\left( { - 1} \right)^{\,n - 1} }} {{2^{\,2n - 1} n}}\frac{{\Gamma (2n - 1)}} {{\Gamma (n)^{\,2} }}x^{\,n} } $$ and then consider the Duplication formula $$ \Gamma \left( {2z} \right) = \frac{{2^{\,2\,z - 1} }} {{\sqrt \pi }}\Gamma \left( z \right)\Gamma \left( {z + 1/2} \right) = \frac{{2^{\,2\,z - 1} }} {{\Gamma \left( {1/2} \right)}}\Gamma \left( z \right)\Gamma \left( {z + 1/2} \right) $$ so that the second term in the summand will become $$ \frac{{\Gamma (2n - 1)}} {{\Gamma (n)^{\,2} }} = \frac{{\Gamma \left( {2\left( {n - 1/2} \right)} \right)}} {{\Gamma (n)^{\,2} }} = \frac{{2^{\,2\,n - 2} }} {{\Gamma \left( {1/2} \right)}}\frac{{\Gamma \left( {n - 1/2} \right)}} {{\Gamma \left( n \right)}} $$ Therefore the whole coefficient of $x^n$ reduces to: $$ \begin{gathered} \frac{{\left( { - 1} \right)^{\,n - 1} }} {{2^{\,2n - 1} n}}\left( \begin{gathered} 2(n - 1) \\ n - 1 \\ \end{gathered} \right) = \frac{{\left( { - 1} \right)^{\,n - 1} }} {{2^{\,2n - 1} n}}\frac{{\Gamma (2n - 1)}} {{\Gamma (n)^{\,2} }} = \hfill \\ = \frac{{\left( { - 1} \right)^{\,n - 1} }} {{2^{\,2n - 1} n}}\frac{{2^{\,2\,n - 2} }} {{\Gamma \left( {1/2} \right)}}\frac{{\Gamma \left( {n - 1/2} \right)}} {{\Gamma \left( n \right)}} = \hfill \\ = \frac{{\left( { - 1} \right)^{\,n - 1} }} {{2n}}\frac{{\Gamma \left( {n - 1/2} \right)}} {{\Gamma \left( {1/2} \right)\Gamma \left( n \right)}} = \frac{{\left( { - 1} \right)^{\,n - 1} }} {{2n}}\left( \begin{gathered} n - 1 - 1/2 \\ n - 1 \\ \end{gathered} \right) = \hfill \\ = \frac{1} {{2n}}\left( \begin{gathered} - 1/2 \\ n - 1 \\ \end{gathered} \right) = \frac{{\frac{1} {2}\Gamma (1/2)}} {{n\;\Gamma (n)\;\Gamma (1 + 1/2 - n)}} = \frac{{\Gamma (1 + 1/2)}} {{\Gamma (1 + n)\;\Gamma (1 + 1/2 - n)}} = \hfill \\ = \left( \begin{gathered} 1/2 \\ n \\ \end{gathered} \right) \hfill \\ \end{gathered} $$

Note that the Duplication formula stems from splitting the product into even and odd components as follows $$ \begin{gathered} \prod\limits_{0\, \leqslant \,k\; \leqslant \,n} {\left( {x + k} \right)} = \prod\limits_{0\, \leqslant \,j\; \leqslant \,\left\lfloor {n/2} \right\rfloor } {\left( {x + 2j} \right)} \;\prod\limits_{0\, \leqslant \,j\; \leqslant \,\left\lfloor {\left( {n - 1} \right)/2} \right\rfloor } {\left( {x + 2j + 1} \right)} = \hfill \\ = 2^{\,\left\lfloor {n/2} \right\rfloor + 1} \prod\limits_{0\, \leqslant \,j\; \leqslant \,\left\lfloor {n/2} \right\rfloor } {\left( {x/2 + j} \right)} \;\;2^{\,\left\lfloor {\left( {n - 1} \right)/2} \right\rfloor + 1} \prod\limits_{0\, \leqslant \,j\; \leqslant \,\left\lfloor {\left( {n - 1} \right)/2} \right\rfloor } {\left( {x/2 + 1/2 + j} \right)} = \hfill \\ = \prod\limits_{0\, \leqslant \,j\; \leqslant \,\left\lfloor {n/2} \right\rfloor } {\left( {x + 2j} \right)\prod\limits_{0\, \leqslant \,j\; \leqslant \,\left\lfloor {\left( {n - 1} \right)/2} \right\rfloor } {\left( {x - 1 + 2\left( {j + 1} \right)} \right)} } = \hfill \\ = \quad \; \cdots \hfill \\ \end{gathered} $$ with all the various "manipulations" you can do on the terms and on the multiplication bounds.