Does a value for $\sqrt{x+\sqrt{x+\sqrt{x+...}}}$ actually exist?

Question

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!

Answer 1

For all $x > 0$, define a sequence $(a_n)$ by $$a_0 = 0 \quad \quad \text{and} \quad a_{n+1} = \sqrt{x + a_n}$$

Let's define the function $$f : t \mapsto \sqrt{x+t}$$

such that $a_{n+1}=f(a_n)$. The function $f$ is increasing on $[0, +\infty)$.

First the sequence is well-defined and bounded. Indeed, one can prove by induction, because $f$ is increasing, that for all $n$, $$0 \leq a_n \leq \frac{1+\sqrt{1+4x}}{2}$$

Now, because the function $f$ is increasing on $[0, +\infty)$, then the sequence $(a_n)$ is monotonous. Because $a_1 = \sqrt{x} > a_0$, one deduces that $(a_n)$ is increasing.

So the sequence is increasing, and bounded, and hence has a limit.

It is natural to define $\sqrt{x+\sqrt{x+\sqrt{x+...}}}$ to be the limit value of this sequence.

Answer 2

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.

Answer 3

Kitchen's Calculus of One Variable gives a rigorous solution although is not exactly the same question (Exercise 8, Section 3-3). To show that $\sqrt{x}$, $\sqrt{x+\sqrt{x}}, \cdots $ converges to $\frac{1+\sqrt{1+4x}}{2}$, let the nth term be $a_n$. Define $h_n=\frac{1+\sqrt{1+4x}}{2}-a_n$. You can prove by induction that $$0<h_n\le \frac{\frac{1+\sqrt{1+4x}}{2}-\sqrt{x}}{(\frac{1+\sqrt{1+4x}}{2}+\sqrt{x})^{n-1}}.$$ Now it is easy to see that $h_n\rightarrow 0$ by pinching since the denominator above is greater than 1 when $x>0$ .