Recall the well-known series
[tex]\displaystyle 2 \arcsin^2(x) = \sum_{n=1}^\infty \frac{(2x)^{2n}}{n^2 \binom{2n}n}[/tex]
Replace x with √x :
[tex]\displaystyle 2 \arcsin^2(\sqrt x) = \sum_{n=1}^\infty \frac{(4x)^n}{n^2 \binom{2n}n}[/tex]
Differentiate both sides:
[tex]\displaystyle -\frac{\arcsin(\sqrt x)}{2\sqrt{x-x^2}} = \sum_{n=1}^\infty \frac{(4x)^n}{n \binom{2n}n}[/tex]
Multiply by 4x :
[tex]\displaystyle -\frac{4x \arcsin(\sqrt x)}{2\sqrt{x-x^2}} = \sum_{n=1}^\infty \frac{(4x)^{n+1}}{n \binom{2n}n}[/tex]
All the series I've mentioned converge for |x| < 1, so we can take x = -1/4 to find the value of the sum we want.
[tex]\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n \binom{2n}n} = -\frac{4 \times \left(-\frac14\right) \arcsin\left(\sqrt{-\frac14}\right)}{2\sqrt{-\frac14-\left(-\frac14\right)^2}} = -\frac{\arcsin\left(\frac i2\right)}{\frac{\sqrt5\,i}2}} = \boxed{\frac2{\sqrt5} \mathrm{arsinh}\left(\frac12\right)}[/tex]
where
[tex]\arcsin\left(\frac i2\right) = z \iff 2\sin(z) = -2i\sinh(iz) = i \implies z = i \, \mathrm{arsinh}\left(\frac12\right)[/tex]
