# Find which initial conditions lead to convergence

## Homework Statement

Let $b_1\in \mathbb{R}$ be given and $n=1,2,\dots$ let $$b_{n+1} := \frac{1+b_n^2}{2}.$$ Define the set $$B := \{b_1\in\mathbb{R} \mid \lim_{n\to\infty}b_n \text{ converges}\}$$

Identify the set $B$.

## The Attempt at a Solution

I claim that $B = [-1,1]$.

First, we note that $\forall b_1 \not = \pm 1$, the sequence is increasing. The base case holds because if $|1-b_1|> 0$ then $\displaystyle \frac{1+b_1^2}{2} > b_1$. Suppose that for some $k$ we have that $b_{k} \ge b_{k-1}$. Then $$b_{k+1} - b_k = \frac{1+b_{k}^2}{2} - \frac{1+b_{k-1}^2}{2}= \frac{(b_k+b_{k-1})(b_l-b_{k-1})}{2},$$ and the latter expression is positive by the inductive hypothesis.

If $b_1 \in (-\infty, 1) \cup (1,\infty)$, then the sequence will diverge to $\infty$, since it is increasing. So suppose that $b_1\in (-1,1)$. Then the sequence is bounded above since $b_{k+1} = \frac{1+b_k^2}{2} < \frac{1+1}{2} = 1$. So by the monotone convergence theorem, the sequence converges in this case. Also, if $b_1=\pm 1$, then $b_n = 1$ for all $n>1$ and so converges to $1$. Hence, $B = [-1,1]$.

I think a direct proof instead of an induction is shorter: $b_{n+1}=\dfrac{1+b_n^2}{2}>b_n \Longleftrightarrow (b_n-1)^2>0$ and done.
The rest is o.k., except for the typo $b_1 \in (-\infty, 1) \cup (1,\infty)\longrightarrow b_1 \in (-\infty, -1) \cup (1,\infty)$