<s>[INST] Infinite product of $\sqrt{2}$ 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?
 [/INST] $$\begin{align}\frac{2}{1}\frac{2}{3} &= 4^1\frac{2!^3}{4!1!^2}\\\frac{2}{1}\frac{2}{3}\frac{6}{5}\frac{6}{7} &= 4^2\frac{4!^3}{8!2!^2}\\\frac{2}{1}\frac{2}{3}\frac{6}{5}\frac{6}{7}\frac{10}{9}\frac{10}{11} &= 4^3\frac{6!^3}{12!3!^2}\\\ldots\end{align}$$
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).
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