Prove that
$$\frac{1}{\sqrt{1 + x} + \sqrt{1 - x}} = \frac{1}{2} + \sum\limits_{k=1}^{\infty}\frac{(4k-1)!!}{(4k+2)!!}x^{2k} , \forall x \in (-1, 1)$$
where $x!!$ means double factorial in this case.
I deducted the following formulae $$(2k)!! = (2k)(2k-2)\cdot\ldots\cdot 2 = 2^k\cdot k!$$ $$(2k+1)!!=(2k+1)(2k-1)(2k-3)\cdot\ldots\cdot 1 = \frac{(2k+1)!}{2^kk!}$$
and tried to replace them in the initial equation but didn't arrive at a final result. I have also tried to rewrite the left-hand side using Taylor on $\frac{\sqrt{1+x}-\sqrt{1-x}}{2x}$.
Your step to write the expression as $\frac{\sqrt{1+x} - \sqrt{1-x}}{2x}$ is correct. You need to expand each expression in the numerator in Generalized binomial series. Keep in mind $$ (1+x)^{\frac{1}{2}} = \sum_{k=0}^{\infty} \binom{\frac{1}{2}}{k}x^k $$ The binomial coefficient can be expanded in the normal way, set $\frac{1}{2} = \alpha$: $$ \binom{\alpha}{k} = \frac{1}{2} (\frac{1}{2} -1) \ldots(\frac{1}{2} - k +1 ) = (-1)^{k}\frac{(2k+1)!!}{2^k} $$ Can you handle from here?