Let \(a>0\) and define \[ x_0=0, \qquad x_{n+1}=a+x_n^2. \] Find a necessary and sufficient condition on \(a\) for the finite limit \[ \lim_{n\to\infty}x_n \] to exist.
Solution
1. If \((x_n)\) has a finite limit, then\[ a\leq\frac14. \]
Suppose \[ x_n\longrightarrow L\in\RR. \] Passing to the limit in \[ x_{n+1}=a+x_n^2 \] gives \[ L=a+L^2. \] Thus \(L\) is a real root of \[ t^2-t+a=0. \] Its discriminant must be nonnegative: \[ 1-4a\geq0. \] Therefore \(a\leq1/4\).
2. Assume\[ 0<a\leq\frac14 \]and set\[ r\coloneqq\frac{1-\sqrt{1-4a}}2. \]Then\[ 0<r\leq\frac12 \qquad\text{and}\qquad a+r^2=r. \]
The displayed value \(r\) is the smaller real root of \[ t^2-t+a=0. \] Since \(0<a\leq1/4\), one has \(0\leq\sqrt{1-4a}<1\), which gives \(0<r\leq1/2\). The root equation is equivalent to \(a+r^2=r\).
3. For every \(n\geq0\),\[ 0\leq x_n\leq r. \]
2. Induction proves the claim.
4. The sequence \((x_n)\) is strictly increasing.
First, \[ x_1=a>0=x_0. \] If \(x_n>x_{n-1}\geq0\), then \[ x_{n+1}-x_n = x_n^2-x_{n-1}^2 = (x_n-x_{n-1})(x_n+x_{n-1}) > 0. \] Induction gives \(x_{n+1}>x_n\) for every \(n\).
5. If \(0<a\leq1/4\), then\[ \lim_{n\to\infty}x_n = r = \frac{1-\sqrt{1-4a}}2. \]
Steps <1>3–<1>4 show that \((x_n)\) is increasing and bounded above, hence it converges to some finite \(L\). As in step 1, \[ L=a+L^2, \] so \(L\) is one of the two roots \[ r = \frac{1-\sqrt{1-4a}}2, \qquad s = \frac{1+\sqrt{1-4a}}2. \] Step 3 gives \(L\leq r\), while \(r\leq s\). Hence \(L=r\).
6. The necessary and sufficient condition is\[ \boxed{0<a\leq\frac14}. \]
Necessity is step 1, and sufficiency is step 5.
7. Q.E.D.
Step 6 is exactly the requested criterion.