New Year Sum

We apply a special case of Kishlaya's identity by letting $\begin{aligned} f(x)=\sum^{\infty}_{k=1}\frac{x^k}{k\dbinom{2k}{k}}=2\sqrt{\frac{x}{4-x}}\arcsin\left(\frac{\sqrt{x}}{2}\right). \end{aligned}$ Differentiating, we obtain $\begin{aligned} f'(x)=\sum^{\infty}_{k=1}\frac{x^{k-1}}{\dbinom{2k}{k}}=\frac{4\arctan\sqrt{\frac{x}{4-x}}}{(4-x)^2\sqrt{\frac{x}{4-x}}}+\frac{1}{4-x}. \end{aligned}$ Now, by substituting $x=-1$ , we get $\begin{aligned} f'(-1)=\sum^{\infty}_{k=1}\frac{(-1)^{k-1}}{\dbinom{2k}{k}}=\frac{4\arctan\sqrt{-\frac{1}{5}}}{5i\sqrt{5}}+\frac{1}{5}. \end{aligned}$ Here comes the interesting bit, how should we evaluate $\arctan\sqrt{-\frac{1}{5}}$ ? Not to worry, we have the inverse hyperbolic tangent function to save the day. Thus, $\begin{aligned} \arctan\sqrt{-\frac{1}{5}}&=&i\tanh^{-1}\sqrt{\frac{1}{5}}\\ &=&\frac{i}{2}\ln\frac{1+\frac{1}{\sqrt{5}}}{1-\frac{1}{\sqrt{5}}}\\ &=&i\ln\frac{1+\sqrt{5}}{2}\\ &=&i\ln\varphi. \end{aligned}$ With that, we obtain the desired result, giving $A\times B=20$ .

To solve hints are:

1) Recognise the sum and write it in form of gamma's.

2) Convert it into form of beta function.

3) Interchange the positions of integral and summation signs.

4) Use Concept of ArithmeticoGeometric Series and convert it all into terms of $x$ .

5) Finally integrate the expression.