Proposition: $\sqrt{x + \sqrt{x + \sqrt{x + ...}}} = \frac{1 + \sqrt{1 + 4x}}{2}$.

Question

Proposition: $$\sqrt{x + \sqrt{x + \sqrt{x + ...}}} = \frac{1 + \sqrt{1 + 4x}}{2}$$

I believe that this is true, and, using Desmos Graphing Calculator, it seems to be true.

I will add how I derived the formula in a moment, if you would like.

Working

I will be honest; all that I did was use the Desmos Graphing Calculator, and let $y = \sqrt{x + \sqrt{x + \sqrt{x + ...}}},$ let $x = 1, 2, 3, ...$, looked at the point at which the two graphs meet, and searched for the number in Google.

It turned up an interesting website, which you may access here, which seemed to show a pattern.

I used this pattern to derive the formula that I stated earlier.

Answer 1

it is true because : $$S=\sqrt { x+\sqrt { x+\sqrt { x+... } } } \\ S=\sqrt { x+S } \\ { S }^{ 2 }-S-x=0\\ S=\frac { 1+\sqrt { 1+4x } }{ 2 } $$

Answer 2

First, I would like to mention that said series is not necessarily well-defined; you can't just add a "..." and expect the resulting quantity to be well-defined. For example, consider: $$S=1+2+4+8+16+\ldots$$ Then, $$2S=2+4+8+16+\ldots$$ so: $$2S+1=S$$ and you get $S=-1$, which is clearly absurd. The issue here is that $S$ is not well defined; you have to define $S$ as the limit as $n \rightarrow +\infty$ of $1+2+\ldots+2^n$, and one can show that this quantity is $+\infty$. Similarly, you must instead ask for the limit of the sequence $a_n$, where $a_0=\sqrt{x}$ and $a_{n+1}=\sqrt{x+a_n}$. Now one can show that this sequence is increasing and is upper bounded, so it must converge (for $x>0$). Only once you have shown that it has converged can you use Nemo's approach (the comments about the quadratic formula's two solutions are easily resolved since the limit has to be positive).