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
$$n_{1,2}=\frac{\sqrt{k}}{2} \pm \sqrt{\frac{k^2}{2}+k}$$

Its strange that there are two solutions, so please check my solution for errors. But the main idea in these kind of questions is always to get back to the first form.
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
$$n_{1,2}=\frac{\sqrt{k}}{2} \pm \sqrt{\frac{k^2}{2}+k}$$

Its strange that there are two solutions, so please check my solution for errors. But the main idea in these kind of questions is always to get back to the first form.