First of all let's assume the series is convergent. Looking for fixed points we have: x=\sqrt{1-\sqrt{1+x}} Now we will try to solve this equation. First squaring both sides: 1-x^2=\sqrt{1+x} \\ \left(\left( 1-x\right)\left( 1+x\right) \right)^2=1+x ...

We may adopt the technique for Ramanujan's infinite radical . Let p(x) = x^2 + 3x + 1 and define F : [0, \infty) \to [0, \infty) by F(x) = \sqrt{p(x) + \sqrt{p(x+1) + \sqrt{p(x+2) + \cdots }}} ...

\displaystyle={0} Explanation: \displaystyle{\left(\sqrt{{2}}\cdot\sqrt{{2}}\right)} + \displaystyle{\left(\sqrt{{2}}\cdot-\sqrt{{2}}\right)} + \displaystyle{\left({0}\cdot{0}\right)} ...

For any n > 0, we have s_n \stackrel{def}{=} \log\left(\sqrt{1!\sqrt{2!\sqrt{3!\sqrt{\cdots\sqrt{n!}}}}}\right) = \sum_{k=1}^n \frac{\log(k!)}{2^k} = \sum_{k=1}^n \frac{1}{2^k}\sum_{j=1}^k\log(j) ...

There's no contradiction, there's just ill-defined terms. The row 3=\sqrt{1+2\cdot \sqrt{1+3 \cdot \sqrt{1+4\cdot \sqrt36}} } is correct, but the following \vdots 3=\sqrt{1+2\cdot \sqrt{1+3 \cdot \sqrt{1+4\cdot \sqrt{1+ \cdots}}} } ...

The axiom you linked to are those of a vector space , not those for a field. It is true (but it is probably not your intention) that F is a vector space with scalar field \mathbb{Q}: F is ...