An uncommon continued fraction of $\frac{\pi}{2}$

Question

I'm currently stuck with the following infinite continued fraction: $$\frac{\pi}{2}=1+\dfrac{1}{1+\dfrac{1\cdot2}{1+\dfrac{2\cdot3}{1+\dfrac{3\cdot 4}{1+\cdots}}}}$$ There is an obscure clue on this: as one can derive the familiar Lord Brouncker’s fraction below $$ \frac{4}{\pi}=1+\dfrac{1^{2}}{2+\dfrac{3^{2}}{2+\dfrac{5^{2}}{2+\dfrac{7^{2}}{2+\cdots}}}} $$ from the Wallis' Formula:

$$ \dfrac{2}{\pi}=\frac{1 \cdot 3}{2 \cdot 2} \cdot \frac{3 \cdot 5}{4 \cdot 4} \cdot \frac{5 \cdot 7}{6 \cdot 6} \cdot \frac{7 \cdot 9}{8 \cdot 8} \cdots $$ the first fraction can be proved in the same manner.

However, I'm not getting any close to it using the Wallis' Formula. Really appreciated if anyone could point me the right direction or explain further how to systematically derive those continued fractions from any given convergent cumulative product.

Answer 1

The result comes from a rather slight modification of Euler's continued fraction , $$a_{0}+a_{0}a_{1}+a_{0}a_{1}a_{2}+....=\cfrac{a_0}{1-\cfrac{a_1}{1+a_1-\cfrac{a_2}{1+a_2-\cfrac{a_3}{\cdots}}}}\label{1}\tag{1}$$

Note that for a product $a_0a_1a_2\cdots$; we can represent them as, $$\begin{align}a_{1}a_{2}a_{3}...&=a_{1}+a_{1}\left(a_{2}-1\right)+a_{1}a_{2}\left(a_{3}-1\right)+\cdots\label{2}\tag{2}\end{align}$$ With \eqref{2} in \eqref{1}, $$a_{1}a_{2}a_{3}a_{4}\cdots=1+\cfrac{a_{1}-1}{1-\cfrac{a_{1}\left(a_{2}-1\right)}{a_{2}a_{1}-1-\cfrac{\left(a_{1}-1\right)a_{2}\left(a_{3}-1\right)}{a_{3}a_{2}-1-\cfrac{\left(a_{2}-1\right)a_{3}\left(a_{4}-1\right)}{\cdots}}}}\label{3}\tag{3}$$

Now apply the Wallis product, $$\frac{\pi}{2}=\frac{2\cdot2}{1\cdot3}\cdot\frac{4\cdot4}{3\cdot5}\cdot\frac{6\cdot6}{5\cdot7}\cdots$$ in \eqref{3} with $a_1=2/1$, $a_2=2/3$ ... to get the first result

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The second result is not obtained via the Wallis product, but from the Leibniz series for $\pi/4$ using \eqref{1}

Answer 2

We know that $$\sin^{-1}x=\int\underbrace{\color{red}{\frac{1}{\sqrt{1-x^2}}}}_{\text{apply binomial theorem}}dx=\int1+\sum_{n=0}^{\infty}\frac{2n-1}{2^{n+1}}x^{2n}dx$$and if you solve this further you'll get $$\sin^{-1}x=x+ \frac{x^3}{6}+\frac{3x^5}{40}+\frac{5x^7}{112}+......$$ $$= x+\frac{1}{2}\cdot\frac{x^3}{3}+\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{x^5}{5}+\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\frac{x^7}{7}+......$$ $$=x+x\left(\frac{x^2}{2\cdot3}\right)+x\left(\frac{x^2}{2\cdot3}\right)\left(\frac{(3x)^2}{4\cdot5}\right)+.........$$ OR $$\sin^{-1}x=x\left(1+\sum_{n=0}^{\infty}\prod_{i=0}^{n}\frac{(2i+1)^2x^2}{(2i+2)\cdot(2i+3)}\right)$$

Now according to Euler's formula for Continued Fraction , i.e. $$S=a\left(1+\sum_{i=1}^{\infty}\prod_{j=1}^{i}r_j\right)=\large\frac{a}{1-\frac{r_1}{1+r_1-\frac{r_2}{1+r_2-\frac{r_3}{......}}}}$$

So , $$\sin^{-1}x=x\left(1+\sum^{\infty}_{n=1}\prod_{i=1}^{n}\frac{(2i-1)^2x^2}{(2i)\cdot(2i+1)}\right)$$ $$=x\Large\left(\frac{1}{1-\frac{\frac{x^2}{2\cdot3}}{1+\frac{x^2}{2\cdot3}-\frac{\frac{3^2x^2}{4\cdot5}}{1+\frac{3^2x^2}{4\cdot5}-.......}}}\right)$$

Since , $$\sin^{-1}1=\frac{\pi}{2}$$ Now just put $x=1$ in the above continued fraction to get to your answer .