A confusing series

Consider the binomial expansion or Taylor's series of $\sqrt{1+x}$ .

$\begin{aligned} (1+x)^\frac 12 & = 1 + \frac 12 x + \frac {\left(\frac 12\right)\left(-\frac 12\right)}{2!}x^2 + \frac {\left(\frac 12\right)\left(-\frac 12\right)\left(-\frac 32\right)}{3!}x^3 + \frac {\left(\frac 12\right)\left(-\frac 12\right)\left(-\frac 32\right)\left(-\frac 52\right)}{4!}x^4 + \cdots \\ & = 1 + \frac 12 x + \sum_{n=2}^\infty \frac {(-1)^{n-1}(2n-3)!!}{2^nn!}x^n \\ & = 1 + \frac 12 x + \sum_{n=2}^\infty \frac {(-1)^{n-1}(2n-2)!}{2^{2n-1}(n-1)!n!}x^n \\ & = 1 + \sum_{n=1}^\infty \frac {(-1)^{n-1}(2n-2)!}{2^{2n-1}(n-1)!n!}x^n \end{aligned}$

Putting $x=1$ , we have $\displaystyle \sum_{n=1}^\infty \frac {(-1)^{n-1}(2n-2)!}{2^{2n-1}(n-1)!n!} = \sqrt 2 - 1 \approx \boxed{0.41421}$ .

This series is the taylor series of f(x)= $\sqrt{x}$ around point a=1 and x=2 without the first term. So the answer is $\sqrt{2}$ - 1 or 0.41421

(This is my first problem so sorry if it's bad.)