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!
 A: For the first product note that, taking $k\geq3
 $ $$\left(1+\frac{1}{k}\right)\left(1-\frac{1}{k+1}\right)=1
 $$ then $$\prod_{k\geq2}\left(1+\frac{\left(-1\right)^{k+1}}{k}\right)=1-\frac{1}{2}=\frac{1}{2}.
 $$ For the second product, which can be written as kennytm suggest, we have $$\prod_{k\geq1}\left(1-\frac{1}{4k-2}\right)\left(1-\frac{1}{4k}\right)\left(1+\frac{1}{2k+1}\right)=\lim_{n\rightarrow\infty}\prod_{k\leq n}\left(1-\frac{1}{4k-2}\right)\left(1-\frac{1}{4k}\right)\left(1+\frac{1}{2k+1}\right)
 $$ and using the identity, for a positive integer $a
 $ $$\Gamma\left(a+\frac{1}{2}\right)=\frac{\sqrt{\pi}\left(2a-1\right)!!}{2^{a}}
 $$ and the functional equation $$\Gamma\left(1+z\right)=z\Gamma\left(z\right)
 $$ we get $$\prod_{k\leq n}\left(1-\frac{1}{4k-2}\right)\left(1-\frac{1}{4k}\right)\left(1+\frac{1}{2k+1}\right)=\frac{\sqrt{\pi}\left(n+1\right)\Gamma\left(2n+\frac{1}{2}\right)}{2^{2n+1}\Gamma\left(n+\frac{1}{2}\right)\Gamma\left(n+\frac{3}{2}\right)}
 $$ hence $$\prod_{k\geq1}\left(1-\frac{1}{4k-2}\right)\left(1-\frac{1}{4k}\right)\left(1+\frac{1}{2k+1}\right)=\lim_{n\rightarrow\infty}\frac{\sqrt{\pi}\left(n+1\right)\Gamma\left(2n+\frac{1}{2}\right)}{2^{2n+1}\Gamma\left(n+\frac{1}{2}\right)\Gamma\left(n+\frac{3}{2}\right)}
 $$ which can be calculated using the Stirling's approximation. Finally $$\prod_{k\geq1}\left(1-\frac{1}{4k-2}\right)\left(1-\frac{1}{4k}\right)\left(1+\frac{1}{2k+1}\right)=\frac{1}{2\sqrt{2}}.
 $$
A: 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
$$
\frac{2}{\pi} = \prod_{k=1}^\infty \left(1-\frac{1}{(2k)^2}\right) \\
\frac{4}{\sqrt{2}\pi} = \prod_{k=1}^\infty\left(1-\frac{1}{(4k)^2}\right) \\
2P \cdot \frac{4}{\sqrt{2}\pi} = \frac{2}{\pi}\\
P = \frac{1}{2\sqrt{2}}
$$
