Recursive square root problem

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  • $\sqrt{c+\sqrt{c+\sqrt{c+\cdots}}}$, or the limit of the sequence $x_{n+1} = \sqrt{c+x_n}$

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The problem asks us to assign a precise meaning to the expression.

Let $a_0=1$, and for every $n\ge 0$, let
$$a_{n+1}=\sqrt{1+a_n}.$$
The precise meaning of the expression is
$$\rho=\lim_{n\to\infty}a_n.$$

Remark: The limit exists, and a version of your argument shows that the limit is indeed $\frac{1+\sqrt{5}}{2}$.

Here is another example of a similar problem. Assign a precise meaning to
$$\rho=1+2+4+8+\cdots.$$

We could (?) say $\rho=1+2\rho$ and therefore (??) $\rho=-1$. It is fairly unlikely (though not impossible) that we would really want to say that $1+2+4+\cdots$ means $-1$.

To “give a precise meaning to” is quite broad. A general approach would be to set
$$\rho_0 = 1; \qquad \rho_{n+1} = \sqrt{1+ \rho_n}$$
Then to show that $(\rho_n)_n$ is convergent (i.e. cauchy in $\mathbb R$) and to define $\rho$ as the limit, using the completeness of $\mathbb R$.
$$\rho = \lim_{n\to\infty} \rho_n$$
You can then prove that
$$\rho = \frac{1+\sqrt5}2$$


Note that $\rho\neq\frac{1-\sqrt5}2$ simply by showing that each $\rho_n \ge 1>\frac12 >\frac{1-\sqrt5}2$
Also, the choice of $\rho_0$ is arbitrary as long as $\rho_0 \ge -1$. It will only affect the rate of convergence. The closer $\rho_0$ is to $\frac{1+\sqrt5}2$, the faster the sequence will converge.