I am struggling to show the following relation $$ \prod_{k=1}^\infty \left(1 - \frac{(-1)^k}{(2k-1)}\right) = \sqrt 2. $$ I have tried to compute the sum $$ \sum_{k=1}^\infty \log \left(1 - \frac{(-1)^k}{2k-1}\right), $$ by using the expansion for $\log(1+x)$, however, I was not able to evaluate the double sum. Furthermore, I tried to square and reorder (although it should not be possible), but haven't quite got the right track.
Could someone give me a hint for this problem?
On
$$\begin{align}\frac{2}{1}\frac{2}{3} &= 4^1\frac{2!^3}{4!1!^2}\\\frac{2}{1}\frac{2}{3}\frac{6}{5}\frac{6}{7} &= 4^2\frac{4!^3}{8!2!^2}\\\frac{2}{1}\frac{2}{3}\frac{6}{5}\frac{6}{7}\frac{10}{9}\frac{10}{11} &= 4^3\frac{6!^3}{12!3!^2}\\\ldots\end{align}$$ Use Stirlings approximation $n! \sim \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$ to establish the limit (the $e$'s and $\pi$'s will factor out).
It follows from the Euler/Weierstrass factorisation for sine: for $x \in \mathbb{R}$
$$\frac{\sin{x}}{x} = \prod_{j \ge 1} \left(1-\frac{x^2}{\pi^2 j^2}\right)$$
Separating the odd even parts we can rewrite your product as:
$$\begin{aligned} \mathcal{P} & = \prod_{k \ge 1 } \left(1 - \frac{(-1)^k}{2k-1}\right) \\& = \frac{1}{2}\prod_{k \ge 0} \bigg(1-\frac{1}{4k-1}\bigg)\bigg(1+\frac{1}{4k+1}\bigg) \\& = \frac{1}{2} \prod_{k \ge 0} \frac{16k^2-4}{16k^2-1} \\& = 2 \prod_{k \ge 1} \frac{16k^2-4}{16k^2-1} \\& = 2 \prod_{k \ge 1}\frac{\bigg(1-(\dfrac{1}{2})^2/k^2\bigg)}{\bigg(1-(\dfrac{1}{4})^2/k^2\bigg)} \\& = 2 \cdot \frac{\sin{\frac{1}{2} \pi}}{\frac{1}{2}\pi} \bigg/\frac{\sin{\frac{1}{4} \pi}}{\frac{1}{4}\pi} \\& = 2 \cdot \frac{1}{\sqrt{2}} \\& = \sqrt{2}. \end{aligned}$$