Answer
When $n \to \infty$ , then we have $|R_n(x)| \leq |\dfrac{ (2n)!}{n! \ (n+1) 2^{(2n+1/2) (2^{n+1}) }}| \to 0$.
Work Step by Step
The remainder in the Taylor series expansion of the function $f(x)$ centred at $a$ can be written as:
$|R_n(x)|=\dfrac{f^{n+1}(c)(x-a)^{n+1}}{(n+1)!}$
Now, we have the Taylor series expansion of the function $f(x)$ centred at $a=0$ as:
$f^{(n+1)}(c)=|\dfrac{(-1)^n (2n)!}{2^{2n+1}(n!) (1+c)^{2n+1/2}}| \leq |\dfrac{ (2n)!}{n! \ 2^{(2n+1/2)}}| $ for $c \in (0,x)$
and $|x|^{n+1} \leq \dfrac{1}{2^{n+1}}$
So, $|R_n(x)| \leq \dfrac{1}{(n+1)!}$
Therefore, when $n \to \infty$ , then we have $|R_n(x)| \leq |\dfrac{ (2n)!}{n! \ (n+1) 2^{(2n+1/2) (2^{n+1}) }}| \to 0$.
