Let $x_1:=a>0$, $x_{n+1} := x_n+\frac{1}{x_n}$ for all $n \in \mathbb{N}$. Determine if $(x_n)$ is convergent or not.

Question

Let $x_1:=a>0$, $x_{n+1} := x_n+\frac{1}{x_n}$ for all $n \in \mathbb{N}$. Determine if $(x_n)$ is convergent or not.

Notice that $x_n>0$ for all $n \in \mathbb{N}$, and $x_n$ is a solution of the equation $t^2-x_{n+1}t+1=0$, and then $\Delta=x^2_{n+1}-4 \geq 0$. So, $x^2_{n+1} \geq 4$.

Thus, $(x_n)$ is bounded below by $4$.

Also we have that

$x_{n+1}-x_n=\frac{1}{x_n}\geq 0$ for all $n\in \mathbb{N}$.

Hence, $(x_n)$ is increasing.

We have shown that $(x_n)$ is increasing and bounded below. It follows by the monotone convergence theorem that $(x_n)$ converges to a limit that is at least $4$, say $x$.

We have $\lim x_{n+1}=\lim (x_n+\frac{1}{x_n})\Rightarrow x=x+\frac{1}{x}$

how can I complete it, please?

If $x = x + \frac{1}{x}$, then $x^2 = x^2 + 1$, whence $1 = 0$; wassup with that? Something seems fishy here . . . — Robert Lewis, Feb 03 '19 at 02:53
Maybe there's a problem with the assertion that $x_n$ is decreasing . . . — Robert Lewis, Feb 03 '19 at 02:56
The monotone convergence theorem does not apply. You should probably re-check your statement of that theorem. — JonathanZ, Feb 03 '19 at 03:34
Additionally, the sequence is not necessarily bounded below by 4, though it is bounded below by $a$. — JonathanZ, Feb 03 '19 at 03:36
@rtybase No, it diverges for any $a>0$. Whether $a=1$ or not, if $L$ is the limit we have $L=L+1/L$. The recurrence in the question at that link is different. — David C. Ullrich, Feb 04 '19 at 13:35

score 2 · Answer 1 · answered Feb 03 '19 at 04:06

2

If there were an integer $N$ such that $x_n\leq N$ for all $n$, then we would have $\tfrac{1}{x_n}\geq \tfrac{1}{N}$ for all $n$. But then $$x_{n+N^2}\geq x_n + N\text{.}$$

answered Feb 03 '19 at 04:06

K B Dave

9,458

score 2 · Answer 2 · answered Feb 03 '19 at 05:03

2

As you have observed the sequence is increasing. either it converges to a finite limit (which is necessarily $>0$) or it converges to $\infty$. The first case is ruled out by taking limits in the given equation. [We cannot have $x=x+\frac 1 x$ ]. Hence $x_n \to \infty$.

answered Feb 03 '19 at 05:03

Kavi Rama Murthy

359,332

Exactly the point I was trying to get at in my comments to the question itself! Cheers! – Robert Lewis Feb 03 '19 at 05:08

Let $x_1:=a>0$, $x_{n+1} := x_n+\frac{1}{x_n}$ for all $n \in \mathbb{N}$. Determine if $(x_n)$ is convergent or not.

2 Answers2