# Does a value for $\sqrt{x+\sqrt{x+\sqrt{x+…}}}$ actually exist? [duplicate]

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!

• The zeta function is a bit different because its an analytical continuation that has applications where jt is correct – Henry Lee Sep 2 '20 at 17:24
• @HenryLee I didn't know that. But that does not mean that $\displaystyle\sum_{n=1}^\infty n = -1/12$, does it? – Rajdeep Sindhu Sep 2 '20 at 17:25
• That js technically the definition of $\zeta(-1)$ but yes, when defined in that context we would just say it diverges – Henry Lee Sep 2 '20 at 17:36
• Well, the real question is, what do you mean by "a value for $\sqrt{x+\sqrt{x+\sqrt{x+...}}}$"? – Eric Wofsey Sep 2 '20 at 20:30
• @EricWofsey For example, $\displaystyle\sum_{n=1}^\infty n$ does not have any value but $\displaystyle\sum_{n=1}^\infty\dfrac{1}{2^n}$ does have one. – Rajdeep Sindhu Sep 3 '20 at 4:37

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.

• "You can prove that this sequence is increasing, and bounded" It would be nice to actually provide a proof. – leonbloy Sep 2 '20 at 17:33
• I agree with @leonbloy – Rajdeep Sindhu Sep 2 '20 at 17:34
• @leonbloy Is that ok now ? – TheSilverDoe Sep 2 '20 at 17:43
• @TheSilverDoe Well, you've proved that the sequence is increasing, but not that is bounded (the induction proof might be easy, I don't know, but I don't see it) – leonbloy Sep 2 '20 at 19:08
• If you are familiar with web diagrams for iterated real functions, the graphs of $f(t) = SQRT(t+x)$ and the line $y=t$ allow you to see immediately that the orbit of $0$ is increasing and bounded (and hence converges to the fixed point). – Ned Sep 2 '20 at 19:46

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.

• I had originally deleted this because it looked similar to TheSilverDoe's answer, but there are some points that I felt could use a better explanation, so I undeleted my answer. – robjohn Sep 2 '20 at 19:05
• Yes it was better at some parts than silver's answer ;) +1 – Baba Yaga Sep 2 '20 at 20:09

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 $$0 Now it is easy to see that $$h_n\rightarrow 0$$ by pinching since the denominator above is greater than 1 when $$x>0$$ .