I would like to find out the exact value (if it can be written down and generalized) for the formula in the title for any x. I have got aproximate values, but I would apreciate an exact value of a special case as well.
I thought about the geometric series and asked myself if there was something for an infinite product as well. After some time I discovered: $$\prod_{n=0}^\infty 1+{1\over x^{2^n} } = {x \over x-1}$$ I found that out by looking at what I had to multiply ${x-1 \over x}$ with to get ${x^2-1 \over x^2}$. With $1+{1\over x^{2^n} } = {x^{2^n}+1 \over x^{2^n}}$ you always use the difference of two squares and it triggers something like a „chain reaction“. Anyway, I tried to find out about: $$\prod_{n=0}^\infty 1+{1\over x^n}$$ I checked with a computer and for values of x bigger than 1 or smaller than -1 it seems to converge. I also plotted the values and they looked similar to ${1\over 1-x}$ but without the part from -1 to 1. After some time I tried to expand it term by term and this pattern emerged: $$\prod_{n=0}^\infty 1+{1\over x^n}=\sum_{n=0}^\infty a(n) {1\over x^n}$$ a(n) here is sequence A122797 in the OEIS. I still don‘t know what this converges to.
I couldn‘t find anything about this on the internet. The only infinite products I could find where with n to some constant power.
I apreciate your help. Even for special cases, like for $x = 2$. Because I don‘t know any exact value of any case. Thanks!