<s>[INST] Does a value for $\sqrt{x+\sqrt{x+\sqrt{x+...}}}$ actually exist? I have seen questions of this type being solved as follows :
$\sqrt{x+\sqrt{x+\sqrt{x+...}}}$'s value does not change if we add an $x$ to the expression and square root it. Let the value of this expression be $y$. So
$$\sqrt{x+y} = y \implies x+y = y^2 \implies y^2-y-x=0$$
Using the quadratic formula, we obtain the value of $y$ as :
$$\dfrac{-(-1)\pm\sqrt{(-1)^2-4(1)(-x)}}{2(1)} = \dfrac{1\pm\sqrt{1+4x}}{2} = \sqrt{x+\sqrt{x+\sqrt{x+...}}}$$
This has been used to solve $\sqrt{6+\sqrt{6+\sqrt{6+...}}}$ in my Mathematics textbook.
Now, this method would work perfectly, assuming that a value for $\sqrt{x+\sqrt{x+\sqrt{x+...}}}$ exists. If a value for this expression does not exist, this would be similar to Numberphile's popular $\displaystyle\sum_{n=1}^\infty n = \dfrac{-1}{12}$ which is doubtlessly wrong.
So, does a value for $\sqrt{x+\sqrt{x+\sqrt{x+...}}}$ exist? Why/why not?
Thanks!
 [/INST] Define the sequence
$$
u_0=0\tag1
$$
$$
u_{n+1}=\sqrt{x+u_n}\tag2
$$
For $x\gt0$, $u_1\gt u_0$, and then by induction and $(2)$, we have
$$
u_{n+1}\gt u_n\tag3
$$
Suppose that $u_n\le\frac{1+\sqrt{1+4x}}2$, then
$$
\begin{align}
u_{n+1}
&\le\sqrt{x+\frac{1+\sqrt{1+4x}}2}\\
&=\frac{1+\sqrt{1+4x}}2\tag4
\end{align}
$$
Since $u_0\le\frac{1+\sqrt{1+4x}}2$, we must have
$$
u_n\le\frac{1+\sqrt{1+4x}}2\tag5
$$
for all $n\ge0$.
Thus, $u_n$ is an increasing sequence that is bounded above. Therefore, the limit exists.
</s>