Infinite Nested Square Root, Prove or disprove that there is at least one real number satisfy

Question

$$\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+...}}}} = 4$$

If I let $X_1=\sqrt x$, $X_2= \sqrt {x+X_1}$, then $X_{n+1}= \sqrt{x+X_n}$ It is a increasing sequence since $X_n<X_{n+1}$ for all $n$. However, is there any way to say it is bounded?

If yes, is my next step to prove the limit to the infinity of the sequence will converge to $4$.

And finally try to figure out the $x$ value? In addition, I saw in Wiki that this infinite square can be represent by a function $$\frac 1 2 (1 + \sqrt {1 + 4n})$$

I am able to follow the logic that when x in side the infinite the square is 2, then $x=\sqrt{2+x}$

However, I don't really understand how it can be generalized to the function shown above.

Answer 1

A way to find an upper bound for the function is to try to anticipate the limit of the sequence, setting $x_n=x_{n+1}=l$..then by induction you can prove that $l$ is an upper bound....

Answer 2

When you are dealing with this nested radicals. the first thing you get is that $x$ should be positive. Now to say that the sequence is bounded, an observation would come in hand, if $x \geq 1$ then by induction each term is less than $2x$. For $x <1$ then this sequence is obviously ,term by term, less than the sequence obtained by $x=1$. So in this case it is again bounded. Now for a sequence which is increasing an bounded in $\Bbb R$ then it shoudld its $\sup$. so the limit exists and if it is eqaul to $L$m then you should solve the quadratic eqaution $ L = \sqrt{x+L}$