# Is there any formula for the infinite surd $\sqrt{k+\sqrt{k^2+\sqrt{k^3+\sqrt{k^4+…}}}}$?

Is there any related studies about the topic?

Let $$n=\sqrt{k+\sqrt{k^2+\sqrt{k^3+\cdots}}}$$ then $$n^2=k+\sqrt{k^2+\sqrt{k^3+\cdots}}$$ By subtracting $$k$$ from both sides and then multiplying out $$\sqrt{k}$$ we get $$n^2 - k=\sqrt{k}\sqrt{k+\sqrt{k^2+\sqrt{k^3+\cdots}}}$$ But now we got the same form as in the beginning, thus we can insert $$n$$ to get $$n^2 - k=\sqrt{k}\cdot n$$ with the solutions $$n_{1,2}=\frac{\sqrt{k}}{2} \pm \sqrt{\frac{k^2}{2}+k}$$
• When you go from equation 2 to equation 3, your $k^2$ does become $k$, but your $k^3$ should also become $k$, while the $k^4$ in the $\dots$ should become $1$. – Todor Markov Dec 4 '19 at 13:43
• You only need two terms to see that this answer is wrong: $\sqrt k\sqrt{k+\sqrt{k^2}}$ is not the same as $\sqrt{k^2+\sqrt{k^3}}$, because $k\sqrt{k^2}\ne\sqrt{k^3}$. – TonyK Dec 4 '19 at 13:44