# Convergence of a recursive sequence

• Mr Davis 97
In summary, the given sequence ##\{a_n\}_{n=1}^\infty## with ##a_1\in\mathbb{N}## does not necessarily converge. This is because if the sequence did converge, it would have to converge to ##L=1## but in the case that ##a_1\geq 2##, the sequence is bounded below by 2 and therefore cannot converge to 1. Thus, the sequence does not converge with ##a_1\in\mathbb{N}##. Additionally, it is shown that ##a_{n+1}>a_n## for all ##n## and ##a_1>1##.
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}##.

## 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##.

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\,.##

fresh_42 said:
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}##.

Mr Davis 97 said:
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\,.##

## 1. What is a recursive sequence?

A recursive sequence is a sequence of numbers in which each term is defined by a formula that depends on one or more of the previous terms. In other words, each term in the sequence is calculated using the terms that came before it.

## 2. How do you determine the convergence of a recursive sequence?

To determine the convergence of a recursive sequence, you must find the limit of the sequence as n approaches infinity. If the limit exists and is a finite number, then the sequence is convergent. If the limit does not exist or is infinite, then the sequence is divergent.

## 3. What is the difference between a convergent and a divergent recursive sequence?

A convergent recursive sequence has a finite limit, meaning that the terms in the sequence eventually get closer and closer to a specific number. On the other hand, a divergent recursive sequence does not have a limit, meaning that the terms in the sequence do not approach a specific number and may grow infinitely larger or smaller.

## 4. Can a recursive sequence have multiple limits?

No, a recursive sequence can only have one limit. If a sequence has multiple limits, it is considered to be oscillating and therefore divergent.

## 5. How do you use the ratio test to determine the convergence of a recursive sequence?

The ratio test is a method for determining the convergence of a recursive sequence. It involves taking the ratio of consecutive terms in the sequence and finding the limit as n approaches infinity. If the limit is less than 1, the sequence is convergent. If the limit is greater than 1, the sequence is divergent. If the limit is equal to 1, the test is inconclusive and further methods must be used to determine convergence.

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