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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 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 this one $$\log 2=\sum_{n=1}^{\infty}\frac{42n-9}{n(2n-1)\binom{4n}{2n}2^{2n+1}}. \tag3$$ One may now ask: can you furnish an alternative proof for the formula in (3)?

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 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)?

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 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).

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 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 this one $$\log 2=\sum_{n=1}^{\infty}\frac{42n-9}{n(2n-1)\binom{4n}{2n}2^{2n+1}}. \tag3$$ One may now ask: can you furnish an alternative proof for the formula in (3)?

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 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.

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 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).