Compute the values of two infinite products whose factors are the same I have the following question:
How to prove that

$(1-\frac{1}{2})\cdot (1+\frac{1}{3})\cdot (1-\frac{1}{4})\cdot (1+\frac{1}{5})\cdot (1-\frac{1}{6})\cdot (1+\frac{1}{7})\cdot (1-\frac{1}{8})\cdot\ ...\ = \frac{1}{2}\ $, but
$(1-\frac{1}{2})\cdot (1-\frac{1}{4})\cdot (1+\frac{1}{3})\cdot (1-\frac{1}{6})\cdot (1-\frac{1}{8})\cdot (1+\frac{1}{5})\cdot (1-\frac{1}{10})\cdot (1-\frac{1}{12})\cdot (1+\frac{1}{7})\cdot\ ...\ = \frac{1}{2 \sqrt{2}}\ ?$

Thanks for the help!

The first is straightforward as already answered by @MarcoCantarini.
For the second
$$P = (1-\tfrac{1}{2})(1-\tfrac{1}{4}) (1+\tfrac{1}{3})(1-\tfrac{1}{6})(1-\tfrac{1}{8}) (1+\tfrac{1}{5}) (1-\tfrac{1}{10}) (1-\tfrac{1}{12}) (1+\tfrac{1}{7}) \cdots
$$
here's an alternate approach. Add a factor $(1+1/1)$ to get
$$
\begin{align}
2P &= (1+\tfrac{1}{1})(1-\tfrac{1}{2})(1-\tfrac{1}{4}) (1+\tfrac{1}{3})(1-\tfrac{1}{6})(1-\tfrac{1}{8}) (1+\tfrac{1}{5}) (1-\tfrac{1}{10}) (1-\tfrac{1}{12}) \cdots \\
&= \prod_{k=0}^{\infty} \left(\frac{2k+2}{2k+1}\right)\left(\frac{4k+1}{4k+2}\right)\left(\frac{4k+3}{4k+4}\right) \\
& = \prod_{k=0}^{\infty} \frac{(4k+1)(4k+3)}{(4k+2)^2} = \frac{1\cdot 3}{2 \cdot 2}\cdot \frac{5 \cdot 7}{6\cdot 6}\cdot \frac{9\cdot 11}{10\cdot 10}\cdots \\
& = \prod_{k=0}^{\infty} \left(1-\frac{1}{(4k+2)^2}\right)
\end{align}
$$
which looks something like the Wallis product for $\pi$, and can be resolved the same way using the infinite product for sine
$$
\frac{\sin(x)}{x} = \prod_{k=1}^\infty \left(1-\frac{x^2}{k^2\pi^2}\right)
$$
Taking $x=\pi/2$ and $x=\pi/4$ we get
$$
\frac{2}{\pi} = \prod_{k=1}^\infty \left(1-\frac{1}{(2k)^2}\right) \\
\frac{4}{\sqrt{2}\pi} = \prod_{k=1}^\infty\left(1-\frac{1}{(4k)^2}\right) \\
2P \cdot \frac{4}{\sqrt{2}\pi} = \frac{2}{\pi}\\
P = \frac{1}{2\sqrt{2}}
$$
