In the following paper, the authors present a proof of:
$$2=(1+1)\left(1+\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{5}\right)\cdot\left(1-\dfrac{1}{7}\right)\cdot\left(1+\dfrac{1}{9}\right)\cdots--(1) (++--++--)$$
(Reference: "New Wallis- and Catalan-Type Infinite Products", by Jonathan Sondow and Huang Yi)
Similarly, by applying the same method, we can prove that: $$\sqrt{2}=(1+1)\left(1-\dfrac{1}{3}\right)\cdot\left(1+\dfrac{1}{5}\right)\cdot\left(1-\dfrac{1}{7}\right)\cdot\left(1+\dfrac{1}{9}\right)\cdots--(2)(+-+-+-+-)$$
This resulted in a generalization for a certain 'n' where n represents the number of alterations of signs (In (1), n=$2$, In (2), n=$1$) as:
$$\prod{\dfrac{\sin(\pi k/(2n))}{\sin(\pi (2k-1)/(4n))}}$$ (k=$1$ to n)
Now I have two questions:
First, How do I calculate the product if 'n' varies. For instance, consider: $(1+1)(1-\dfrac{1}{3})(1+\dfrac{1}{5})(1+\dfrac{1}{7})(1-\dfrac{1}{9})(1-\dfrac{1}{11})\cdots-->(3)(+-++--+++---\cdots)$
How can one evaluate such a product?
Secondly, I used Euler's reflection formula and Weierestrass infinite product of gamma function(the referenced paper's method) to derive the general result. These are admittedly hard concepts for me currently. Can this result be derived in another simpler way perhaps?
Edit: I have found a "solution" for the sign fluctuating with increasing frequency$(3)(+-++--+++---\cdots)$
The product is a variant of $(2)$.
$(3)=\sqrt{2}\cdot\dfrac{1+(1/7)}{1-(1/7)}\cdot\dfrac{1-(1/9)}{1+(1/9)}\cdot\dfrac{1+(1/15)}{1-(1/15)}\cdot\dfrac{1-(1/21)}{1+(1/21)}\cdots$
This results in:
$(3)=\dfrac{4\sqrt{2}}{3}\cdot \left(\dfrac{8\cdot16\cdot20\cdot28\cdot32\cdots}{10\cdot14\cdot22\cdot26\cdot34\cdots}\right)=\dfrac{4\sqrt{2}}{3}\cdot \left(\dfrac{4\cdot8\cdot10\cdot14\cdot16\cdots}{5\cdot7\cdot11\cdot13\cdot17\cdots}\right)=\left(\dfrac{4\sqrt{2}}{3}\cdot \prod{\dfrac{(6n+4)\cdot(6n+8)}{(6n+5)\cdot(6n+7)}}\right)$
$(3)=\dfrac{4\sqrt{2}}{3}\cdot\dfrac{\Gamma{(5/6)}\Gamma{(7/6)}}{\Gamma{(2/3)}\Gamma{(4/3)}}=\dfrac{2\sqrt{2}}{\sqrt{3}}$
(Consequently, from the cosine factorisation, the series$(4)(+--++---+++\cdots)$:$(1+\dfrac{1}{3})(1-\dfrac{1}{5})(1-\dfrac{1}{7})(1+\dfrac{1}{9})(1+\dfrac{1}{11})\cdots=\dfrac{\pi}{4}\cdot\dfrac{2\sqrt{2}}{3}\cdot \dfrac{\Gamma{(2/3)}\Gamma{(4/3)}}{\Gamma{(5/6)}\Gamma{(7/6)}}=\dfrac{\pi\sqrt{3}}{4\sqrt{2}}$
So now my question becomes
1): Can we generalize a formula for any fluctuating sign series?(Could be AP or GP) Or at least find all the possible fluctuations of 'n' for which the final result is deducible.
2): Now, there does exist a simpler proof for the first product, https://math.stackexchange.com/q/4777515. Please try to prove $(1)$ using any method except the one mentioned here and the cited answer. You are free to use advanced mathematics.
This is not an answer but might help in proving.
Proving $(1),1=\left(1+\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{5}\right)\cdot\left(1-\dfrac{1}{7}\right)\cdot\left(1+\dfrac{1}{9}\right)\cdots$ is equivalent to proving: $$\left(1-\dfrac{1}{3}\right)\cdot\left(1+\dfrac{1}{5}\right)\cdot\left(1+\dfrac{1}{7}\right)\cdot\left(1-\dfrac{1}{9}\right)\cdots=\dfrac{\pi}{4}--Conv(1)$$
Since, by multiplying these two expressions, we get the Wallis' product of $\pi$ which states that:
$$\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{5^2}\right)\left(1-\dfrac{1}{7^2}\right)\left(1-\dfrac{1}{9^2}\right)\cdots=\dfrac{\pi}{4}$$
Now if we multiply the reciprocal of $Conv(1)$ and the Wallis' product, we get:
$$\left(\dfrac{2\cdot4\cdot4\cdot6\cdot6\cdot8\cdot8\cdot10\cdot10\cdots}{3\cdot3\cdot5\cdot5\cdot7\cdot7\cdot9\cdot9\cdot11\cdots}\right)\cdot\left(\dfrac{3\cdot5\cdot7\cdot9\cdot11\cdot13\cdot15\cdots}{2\cdot6\cdot8\cdot8\cdot10\cdot14\cdot16\cdots}\right)$$(We have to prove that this equals $1$)