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][1] 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)?

[1]: https://en.wikipedia.org/wiki/Wilf%E2%80%93Zeilberger_pair