In a comment to this question here, I made, based on my "defective" memory, the claim that
\begin{equation} \prod_{n=0}^{\infty}\left(1+ \frac{1}{2^n}\right)=\frac{2}{1-P}, \end{equation} where $P$ is the Pell constant defined as:
\begin{equation} P=1-\prod_{n=0}^\infty\left(1-\frac1{2^{2n+1}}\right) \end{equation}
and, luckily, got corrected by user mrtaurho. However, although my underlying assumption was false, the result still appears to be correct (by a numerical check).
As I don't believe in coincidences but find this relation interesting, I thought I'd ask whether there is anybody here who could prove or disprove this relation?
Thank you very much in advance for your interest.
This is pretty easy to see, if you ignore matters of convergence (but everything converges absolutely anyways so it's fine):
We just want to prove that $$\prod_{m=0}^\infty\left(1+\frac1{2^m}\right)\prod_{n=0}^\infty\left(1-\frac1{2^{2n+1}}\right)=2.$$
which is just
$$\prod_{m=1}^\infty\left(1+\frac1{2^m}\right)\prod_{n=0}^\infty\left(1-\frac1{2^{2n+1}}\right)=1.$$
We match up terms -- a single $1-\frac1 {2^{2n+1}}$ term gets paired with all $m$ such that $2n+1|m$. In other words, we arrange the product to:
$$\prod_{n=0}^\infty\left(\left(1-\frac1{2^{2n+1}}\right)\prod_{m=0}^\infty\left(1+\frac1{2^{2^m(2n+1)}}\right)\right)=1.$$
Since every factor in the infinite product on the left is just $1$, the two sides match like we wanted.