Convergence of a recursive sequence

[Mr Davis 97]

Homework Statement

With $a_1\in\mathbb{N}$ given, define $\displaystyle {\{a_n\}_{n=1}^\infty}\subset\mathbb{R}$ by $\displaystyle {a_{n+1}:=\frac{1+a_n^2}{2}}$, for all $n\in\mathbb{N}$.

Homework Equations

The Attempt at a Solution

We claim that with $a_1 \in \mathbb{N}$, the sequence does not necessarily converge. First, we note that if the sequence did converge to $L$, then $L = \frac{L^2+1}{2} \implies (L-1)^2=0 \implies L=1$ (we used the fact that since $(a_{n+1})$ is a subsequence of $(a_n)$, $\lim a_n = \lim a_{n+1} = L$). We now split the proof up into two cases:Case 1: $a_1=1$. In this case, for all positive integers $n$, $a_n = 1$. Hence $\lim a_n = 1$.

Case 2: $a_1 \ge 2$. First, we show that $(a_n)$ is bounded below by 2. That is, we want to show that $\forall n \in \mathbb{N}$, $a_n \ge 2$. We proceed by induction. The base case clearly holds. Suppose that for some $k \in \mathbb{N}$ we have $a_k \ge 2$. Then $a_{k+1} = \frac{a_k^2+1}{2} \ge \frac{4+1}{2} = \frac{5}{2} \ge 2$. So the sequence is bounded below by $2$. Hence, the sequence does not converge, for if it did, the terms of the sequence would get arbitrarily close to $1$ for large $n$.

 

[fresh_42]

The sequence $(3,3,3,3,3, \ldots)$ is bounded from below by $2$, yet it converges. You should show $a_{n+1}>a_n$ for all $n$ and $a_1>1\,.$

 

[Mr Davis 97]

The sequence $(3,3,3,3,3, \ldots)$ is bounded from below by $2$, yet it converges. You should show $a_{n+1}>a_n$ for all $n$ and $a_1>1\,.$

But the sequence $(3,3,3,3, \dots)$ does not converge to $1$. What I thought I showed was that if the sequence did converge, it would have to converge to $1$, and that in the case that $a_1 \ge 2$, the sequence is bounded below by 2, and so couldn't possibly converge to $1$. Hence it doesn't necessarily converge with $a_1 \in \mathbb{N}$.

 

[fresh_42]

But the sequence $(3,3,3,3, \dots)$ does not converge to $1$. What I thought I showed was that if the sequence did converge, it would have to converge to $1$, and that in the case that $a_1 \ge 2$, the sequence is bounded below by 2, and so couldn't possibly converge to $1$. Hence it doesn't necessarily converge with $a_1 \in \mathbb{N}$.

You're right. I forgot this. Btw., the same argument as for $L=1$ also shows $a_{n+1}>a_n\,.$