Convergence of a recursive sequence

  • #1
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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##.
 

Answers and Replies

  • #2
14,398
11,712
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\,.##
 
  • #3
1,462
44
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}##.
 
  • #4
14,398
11,712
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\,.##
 

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