TITLE: $\sqrt{c+\sqrt{c+\sqrt{c+\cdots}}}$, or the limit of the sequence $x_{n+1} = \sqrt{c+x_n}$
QUESTION [82 upvotes]: (Fitzpatrick Advanced Calculus 2e, Sec. 2.4 #12)
For $c \gt 0$, consider the quadratic equation
$x^2 - x - c = 0, x > 0$.
Define the sequence $\{x_n\}$ recursively by fixing $|x_1| \lt c$ and then, if $n$ is an index for which $x_n$ has been defined, defining
$$x_{n+1} = \sqrt{c+x_n}$$
Prove that the sequence $\{x_n\}$ converges monotonically to the solution of the above equation.
Note: The answers below might assume $x_1 \gt 0$, but they still work, as we have $x_3 \gt 0$.

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REPLY [4 votes]: Hint on methodology:
Denoting $f(x)=\sqrt{c+x}$ the function that defines the recursive sequence, you can show the sequence is increasing (resp.decreasing) by showing that all its terms live in an interval $I$ such that
$$f(x)>x \quad(\text{resp.}\;f(x)<x)\quad\forall x\in I. $$
To show that  all terms of the sequence live in $I$, you just have to show that $f(I)\subseteq I$.
Last point: all terms are obviously positive. On the other hand, as $f$ is continuous, if a limit $\ell$ exists, it satisfies the equation
$$f(x)=\sqrt{c+x}=x\iff x^2-x-c=0,\quad x>0$$
This quadratic equation has a negative and a positive root: $\;\dfrac{1\pm \sqrt{1+4c}}2$. Therefore, if the sequence converges, its limit  is the positive root $\lambda$, and you can try to show that
$$f\bigl([0,\lambda]\bigr)\subseteq[0,\lambda]\quad\text{and}\quad f(x)>x\enspace\text{on }\; [0,\lambda].$$