A Last question on nested radicals but this time with complex value :
$${\Re}\Big(\sqrt{1+\frac{i}{2}\sqrt{1+\frac{i}{2^2}\sqrt{1+\frac{i}{2^3}\sqrt{1+\frac{i}{2^4}\sqrt{\cdots}}}}}\Big)=1$$
I have tried to use power series like in the answer of Somos but I fail.
Furthermore we have the beautiful relation $$S={\Re}\Big(\sqrt{1+\frac{i}{2}\sqrt{1+\frac{i}{2^2}\sqrt{1+\frac{i}{2^3}\sqrt{1+\frac{i}{2^4}\sqrt{\cdots}}}}}\Big)+{\Im}\Big(\sqrt{1+\frac{i}{2}\sqrt{1+\frac{i}{2^2}\sqrt{1+\frac{i}{2^3}\sqrt{1+\frac{i}{2^4}\sqrt{\cdots}}}}}\Big)=\sqrt{1+\frac{1}{2}\sqrt{1+\frac{1}{2^2}\sqrt{1+\frac{1}{2^3}\sqrt{1+\frac{1}{2^4}\sqrt{\cdots}}}}}=1.25$$ Furthermore I think it's good to know what happend in the case of complex number with the Herschfeld's theorem .If somebody have some paper on this it would be cool .
My question :How solve it ?
Thanks in advance .
Looking at the referenced answer by Somos, you want (the real part of) $f(\frac i2,\frac12)$. But if $t\mapsto f(t,\frac12)$ is a power series that converged in a neighbourhood of $0$ and that equals $1+\frac t2$ for infinitely many values of $t$ (see equation $(4)$ in the referenced answer) accumulating at $0$, then by the identitiy theorem, $f(t,\frac12)=1+\frac t2$ holds for all $t$ (where we have convergence). Consequently, $$ \Re\left(f(\tfrac i2,\tfrac12)\right)=\Re(1+\tfrac i4)=1.$$