Question
Show that \sqrt{n-\sqrt{n-\sqrt{n-\sqrt{n-\cdots \infty } } } } = \dfrac{\sqrt{1+4n} -1 }{2}
Answer
Let
x =\sqrt{n-\sqrt{n-\sqrt{n-\sqrt{n-\cdots \infty } } } }
when the nested square roots go to infinite
which makes no difference starting to count from the second radical.
Then, we get the following equation
x = \sqrt{n-x}
Then,
x > 0
Square both sides results in a quadratic equation
x^2+x-n=0
Using the root formula for a quadratic equation, we get
x = \dfrac{-1\pm\sqrt{1^2-4\cdot 1\cdot (-n)} }{2} = \dfrac{-1\pm\sqrt{1+4n} }{2}
Cancel the negative solution, then,
x = \dfrac{\sqrt{1+4n} -1 }{2}
Therfore,
\sqrt{n-\sqrt{n-\sqrt{n-\sqrt{n-\cdots \infty } } } } = \dfrac{\sqrt{1+4n} -1 }{2}