Question

$\displaystyle \sf \color {gold}{ \sum_{n = 1}^{ \infty } \frac{( - 1 {)}^{n + 1} }{ n\binom{2n}{n} } }$​

Answer 1

Answer:

Question-

$\displaystyle \sf \color {red}{ \sum_{n = 1}^{ \infty } \frac{( - 1 {)}^{n + 1} }{ n\binom{2n}{n} } }$

Answer-

$\displaystyle \lim_{n \to + \infty } \frac{1}{n \times \binom{2n } {n}}$

$\displaystyle \lim_{n \to + \infty }(n \times \frac{2n}{n} )$

Evaluate the limit.

= 1

$\therefore \: 1 \\ + \infty$

Since the expression \frac{a}{±∞} , a \in R is defined as 0. The limit

$\quad \bold {Since \: the \: expression \: \frac{a}{±∞} , a \in R \: is \: defined \: as \: 0. \: The \: limit \:}$

$\displaystyle\therefore \lim_{n\to+\infty }( \frac{1}{n \times \binom{2n} {n}}) = 0$

After checking inequality,

$\quad \displaystyle \bold \green{The \: series \: converges \: .}$

Answer 2

Recall the well-known series

$\displaystyle \rm2 \arcsin^2(x) = \sum_{n=1}^\infty \frac{(2x)^{2n}}{n^2 \binom{2n}n}$

Replace x with √x :

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

Differentiate both sides:

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

Multiply by 4x :

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

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.

$\displaystyle \rm\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n \binom{2n}{n} } = -\frac{4 \times \left(-\frac14\right) \arcsin\left(\sqrt{-\frac14}\right)}{2\sqrt{-\frac14-\left(-\frac14\right)^2}} = - \frac{\arcsin\left(\frac i2\right)}{\frac{\sqrt5\,i}2} = \boxed{\frac2{\sqrt5} \mathrm{arsinh}\left(\frac12\right)}$

where

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