# How to solve $y = \sqrt{k\sqrt{k\sqrt{k \sqrt{k\sqrt{\dots}}}}}$?

$\text{Given}\quad y = \sqrt{k\sqrt{k\sqrt{k \sqrt{k\sqrt{\dots}}}}}\quad \text{ where }\,k\geq 0,\;\;\text{find the value of }\,y.$
I have no idea on how to solve problems like this.

## 2 Answers

$$y=\sqrt{k\cdot\sqrt{k\cdot\sqrt{\dots}}}=k^{1/2+1/4+1/8+...}=k^{\frac{1}{2}\frac{1}{1-1/2}}=k^1=k$$

• That's a nice approach too. I hadn't thought of it that way. Dec 30, 2013 at 14:28

We know that $$y=\sqrt{k\cdot\sqrt{k\cdot\sqrt{\dots}}}$$ We can square the equation and divide by $k$: $$\frac{y^2}k=\sqrt{k\cdot\sqrt{\dots}}$$ But the right hand side is just $y$ again, so we have $$\frac{y^2}k=y$$ Solving this gives $y=0$ or $y=k$.

• This answer assumes that $y$ exists. One also needs to show that the sequence of iterated square roots converge. Aug 15, 2014 at 12:46