Question

`\rm \sum_(n = 1)^( \infty ) \frac{( - {1)}^(n + 1) }{n \binom{2n}{n} } \\`​

Answer 1

Recall the well-known series

`\displaystyle 2 \arcsin^2(x) = \sum_(n=1)^\infty \frac{(2x)^(2n)}{n^2 \binom{2n}n}`

Replace x with √x :

`\displaystyle 2 \arcsin^2(\sqrt x) = \sum_(n=1)^\infty \frac{(4x)^n}{n^2 \binom{2n}n}`

Differentiate both sides:

`\displaystyle -(\arcsin(\sqrt x))/(2√(x-x^2)) = \sum_(n=1)^\infty \frac{(4x)^n}{n \binom{2n}n}`

Multiply by 4x :

`\displaystyle -(4x \arcsin(\sqrt x))/(2√(x-x^2)) = \sum_(n=1)^\infty \frac{(4x)^(n+1)}{n \binom{2n}n}`

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.

`\displaystyle \sum_(n=1)^\infty \frac{(-1)^(n+1)}{n \binom{2n}n} = -(4 * \left(-\frac14\right) \arcsin\left(√(-\frac14)\right))/(2√(-\frac14-\left(-\frac14\right)^2)) = -\frac{\arcsin\left(\frac i2\right)}{\frac{\sqrt5\,i}2}} = \boxed{\frac2{\sqrt5} \mathrm{arsinh}\left(\frac12\right)}`

where

`\arcsin\left(\frac i2\right) = z \iff 2\sin(z) = -2i\sinh(iz) = i \implies z = i \, \mathrm{arsinh}\left(\frac12\right)`