$\prod\limits_{k=0 \atop 2k+1\ne 0 \bmod n}^\infty \frac{2k+1+(-1)^{\lfloor (2k+1)/n\rfloor}}{2k+1}=\sqrt{n}$ for $n\ge 2$ I found tasks in an old script. For the following problem I have no idea how to solve.
How can one prove $$\prod\limits_{k=0 \atop 2k+1\ne 0 \bmod n}^\infty \frac{2k+1+(-1)^{\lfloor (2k+1)/n\rfloor}}{2k+1}=\sqrt{n}$$ for $n\in\mathbb{N}$ with $n\ge 2$?   
It generalizes
How to prove that $\prod_{n=0}^\infty \frac{(4n+2)^2}{(4n+1)(4n+3)}=\sqrt{2}$ .
Examples $n=2$ and $n=3$: 
$$\prod\limits_{k=0}^\infty \frac{2k+1+(-1)^k}{2k+1}= \frac{2}{1}\frac{2}{3}\frac{6}{5}\frac{6}{7}\frac{10}{9}\frac{10}{11}…=\sqrt{2}$$ 
$$\prod\limits_{k=0 \atop 2k+1\ne 0 \bmod 3}^\infty \frac{2k+1+(-1)^{\lfloor (2k+1)/3\rfloor}}{2k+1}= \frac{2}{1}\frac{4}{5}\frac{8}{7}\frac{10}{11}\frac{14}{13}\frac{16}{17}…=\sqrt{3}$$ 

Literature: Sierpinski Oeuvres Choisies I (page 222-226) 
 http://www.plouffe.fr/simon/math/Sierpinski%20Oeuvres%20Choisies%20I.pdf 
Gamma function product: $\displaystyle \Gamma(1+x)=\lim\limits_{m\to\infty} m^x/\prod\limits_{k=1}^m \left(1+\frac{x}{k}\right)$ 
or written äquivalent  $\displaystyle m^x/\prod\limits_{k=1}^m \left(1+\frac{x}{k}\right) \sim\Gamma(1+x)$ .
Therefore one get 
$(1)$ $\displaystyle m^x\prod\limits_{k=1}^m \frac{k-x}{k}\sim\Gamma(1-x)$ 
and general  
$(2)$ $\displaystyle (nm)^x\prod\limits_{k=1}^{nm} \frac{k-x}{k}\sim\Gamma(1-x)$
for $n\in\mathbb{N}$ and $x\in\mathbb{R}\setminus\mathbb{N}$. 
It’s $\dfrac{nk}{nk-x}\dfrac{k-x}{k} =1+\dfrac{(1-n)x}{nk-x}$ and with $\dfrac{(2)}{(1)}$ follows therefore 
$$ \prod\limits_{k=1}^m\frac{k-x}{k} \prod\limits_{k=1}^{nm}\frac{k}{k-x}     =  
\prod\limits_{k=1}^m \left(\frac{nk}{nk-x}\frac{k-x}{k}\right) \prod\limits_{ k=1 \atop k\ne 0 \bmod n }^{nm}\left(1+\frac{x}{k-x}\right)  $$ $$=
\prod\limits_{k=1}^{nm}\left(1+\frac{x(1-0^{k-n\lfloor\frac{k}{n}\rfloor}n)}{k-x}\right) \sim n^x$$ 
with $0^0:=1$.  

With $x:=\frac{1}{2}$ the formula of the question is proofed.
