The sum is: $$\frac12-\frac{1\cdot3}{2\cdot4\cdot6}+\frac{1\cdot3\cdot5\cdot7}{2\cdot4\cdot6\cdot8\cdot10}-\frac{1\cdot3\cdot5\cdot7\cdot9\cdot11}{2\cdot4\cdot6\cdot8\cdot10\cdot12\cdot14}+-\dots$$
**background:** so the previous problem was the same question except that each minus sign was replaced with a plus sign, and for that case I managed to solve it by noticing that:$$(1-x)^{-1/2}=1+\frac{x}2+\frac{1\cdot3}{2\cdot4}x^2+\dots\text{ and }(1+x)^{1/2}=1-\frac{x}2+\frac{1\cdot3}{2\cdot4}x^2-\dots$$
Thus summing both sides after substituing $x^2$ for $x$ and multiplying in turn by $x$ one obtains:$$x+\frac{1\cdot3}{2\cdot4}x^5+\frac{1\cdot3\cdot5\cdot7}{2\cdot4\cdot6\cdot8}x^9+\dots=\frac12x[(1+x^2)^{-1/2}+(1-x^2)^{-1/2}]$$from which by integrating and putting $x=1$ one finally obtains the requested sum.
The problem with the alternating one is that this simple trick is not possible here. What do you suggest?
I should be able to do it without complex series (why I mention complex series? because it might be tempting to follow a similar reasoning [identity for sum of binomial coefficients](https://math.stackexchange.com/questions/3698077/identity-for-sum-of-binomial-coefficients))