We all know the series expansion
$$\log 2=\sum\_{n=1}^{\infty}\frac{(-1)^{n-1}}n. \tag1$$
I also am able to use the method of [Wilf-Zeilberger](https://en.wikipedia.org/wiki/Wilf%E2%80%93Zeilberger_pair) to the effect that
$$\log 2=3\sum\_{n=1}^{\infty}\frac{(-1)^{n-1}}{n\binom{2n}n2^n}. \tag2$$



> 
> **QUESTION.** Can you provide yet another proof of the formula in (2)?
> 
> 
> 


**Remark.** My motivation for this question goes beyond this particular series, hoping it paves a way forward in my study.


**Postscript.** After those generous replies (see below), it appears that the idea rests on
$$\log\left(1+\frac1x\right)=2\sinh^{-1}\left(\frac1{2\sqrt{x+x^2}}\right)$$
so that we may put $x=1$ to obtain (1) and (2).


To reveal the background: (2) is found from (1) by a "series acceleration" method which does not even stop there. In fact, stare at these two
\begin{align\*}\log 2&=3\sum\_{n=1}^{\infty}\frac{14n-3}{\binom{2n}2\binom{4n}{2n}2^{2n+1}}, \tag3 \\
\log 2&=3\sum\_{n=1}^{\infty}
\frac{(171n^2 - 111n + 14)(-1)^{n-1}}{\binom{3n}3\binom{6n}{3n}2^{3n+1}} 
\tag4 \end{align\*}
One may now ask: can you furnish an alternative proof for the formulae (3) or (4)?



In search of an alternative proof of a series expansion for $\log 2$
Since you wish to develop techniques, you might want to consider the more general form
$$S\_k=\sum\_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^k\binom{2n}n2^n}.$$
The arcsine representation
$$\arcsin^2z=\frac12\sum\_{n=1}^\infty\frac{(2z)^{2n}}{n^2{2n \choose n}}$$
directly gives
$$S\_2=\tfrac{1}{2}\ln^2 2,$$
(substitute $z=2^{-3/2}i$), upon differentiation one finds
$$S\_1=\tfrac{1}{3}\ln 2,$$
$$S\_0=\tfrac{1}{9}+\tfrac{4}{27}\ln 2,$$
and upon integration,
$$S\_3=\tfrac{1}{4}\zeta (3)-\tfrac{1}{6}\ln^3 2 ,$$
$$S\_4=4\operatorname{Li}\_4\left(\tfrac12\right)-\tfrac72\zeta(4)+\tfrac{13}4\ln2\zeta(3)-\ln^22\zeta(2)+\tfrac5{24}\ln^42.$$
This method apparently fails to give a closed form expression for $k>4$, see this [MSE posting](https://math.stackexchange.com/q/3307589/87355).


