Recursive sequence does not converge

Question

Define $x_n$ recursively as follows: $x_1=1$, $x_{n+1}=x_n+\frac{1}{x_n}$. We are asked to show this sequence is not convergent. Here's my attempt.

Since $x_1=1>0$, and for each $n \in \mathbb{N}, |x_n|>0$, we must have $|\frac{1}{x_n}|>0$ and hence $|x_n|<|x_{n+1}|$. Which means $\{x_n\}$ is a monotone sequence and to show it doesn't converege, it is sufficient to show that it isn't bounded. Let us assume $\lbrace x_n \rbrace$ is bounded and let $M=\sup \lbrace x_n \rbrace$.

By supremum property of $\mathbb{R}$, given $\epsilon> \frac{1}{M} >0, \exists k \in \mathbb{N}, x_k>M-\epsilon$. Say $x_k=\delta>M-\frac{1}{M}$. Note that $M \geq \delta \implies \frac{1}{\delta} \geq \frac{1}{M}$. Also, $M-\frac{1}{M}< \delta \leq M$ $\implies$ $M < \delta +\frac{1}{M} \leq M+\frac{1}{M}$, and from this we get $M< \delta+\frac{1}{M}< \delta+\frac{1}{\delta}$. But, $x_{k+1}=\delta+\frac{1}{\delta}$, which is a contradiction, hence the sequence is unbounded.

Does this look okay?

Edit: Changed the definition of M as per the comment.

$\sup { y | y \ge x_n }$ will always be $+\infty$? Is that what you meant? — copper.hat, May 04 '20 at 18:02
See https://math.stackexchange.com/q/2758372/42969 for simpler solutions. — Martin R, May 04 '20 at 18:06
No, I mean your definition means $M=\infty$ regardless of $x_n$. — copper.hat, May 04 '20 at 18:11

score 3 · Accepted Answer · answered May 04 '20 at 18:13

3

The idea is fine, but some of the details are wrong. First, by your definition $\sup\{y\in\Bbb R:y\ge x_n\}=+\infty$, which definitely isn’t what you want; from what follows it appears that you want to define $M$ to be the assumed least upper bound of the sequence, i.e., $M=\sup\{x_n:n\in\Bbb Z^+\}$.

Your $\epsilon$ is never actually used. Moreover, if $\epsilon>\frac1M$, choosing $k\in\Bbb Z^+$ so that $x_k>M-\epsilon$ does not ensure that $x_k>M-\frac1M$, since $M-\frac1M>M-\epsilon$. Just observe that the definition of $M$ ensures that there is a $k\in\Bbb Z^+$ such that $x_k>M-\frac1M$. There is no need to rename it $\delta$: that just introduces an unnecessary symbol. Then you can complete the argument roughly as you did, but a bit more efficiently: $x_k<M$, so $\frac1{x_k}>\frac1M$, and therefore

$$x_{k+1}=x_k+\frac1{x_k}>\left(M-\frac1M\right)+\frac1M=M\;,$$

contradicting the definition of $M$.

answered May 04 '20 at 18:13

Brian M. Scott

631,399

4

You know, in my book downvoting an answer that answers the OP’s request by showing exactly where the proposed argument went astray and how to correct it is vandalism pure and simple. – Brian M. Scott May 04 '20 at 18:40
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This is such an amazing and informative answer. To be fair, I also dont get why would someone downvote the question as well. The OP is showing effort in understanding. Mind blowing. – James May 04 '20 at 19:52
Your answer was extremely helpful. I sat with my face in my palm for about a minute after reading it, made the proposed edit and green ticked your answer. – Avijit Dikey May 05 '20 at 05:01
@AvijitDikey: I’m glad that it helped. – Brian M. Scott May 05 '20 at 05:38

score 0 · Answer 2 · 2020-05-04T18:27:09.577

0

Assuming that the sequence converges, it converges to some $x_\infty$ such that

$$x_\infty=x_\infty+\frac1{x_\infty}.$$

This proves that convergence is impossible.

edited May 04 '20 at 18:27

answered May 04 '20 at 18:20

Recursive sequence does not converge

2 Answers2