Let $x_1=a>0$ and $x_{n+1}=x_n+\frac{1}{x_n} \forall n\in \mathbb N$. Check whether the following sequence converges or diverges.

Question

Let $x_1=a>0$ and $x_{n+1}=x_n+\frac{1}{x_n} \forall n\in \mathbb N$. Check whether the following sequence converges or diverges.

When I was in UG my teacher used derivative test for monotonicity. $f(x)=x+\frac{1}{x}, f'(x)=1-\frac{1}{x^2}>0(x>1).$ So, $f(x)$ is increasing. How to prove the sequence is monotonic? Differentiation is coming after the sequences and series. By AM-GM inequality sequence is bounded below. $x_{n+1}=x_n+\frac{1}{x_n}\ge 2\sqrt{x_n.\frac{1}{x_n}}=2 \forall n\in \mathbb N$. How can I judge whether the sequence bounded above or not? Please help me.

Answer 1

Rearranging we have $x_{n+1}-x_n=\frac{1}{x_n}$. Multiplying by $x_{n+1}+x_n$ we get $$x^2_{n+1}-x^2_n=1+\frac{x_{n+1}}{x_n} \geq 1$$ Telescoping the sum we obtain that, $$x^2_{n+1}-x_1^2\geq n$$ Hence $x_n$ is unbounded.

Answer 2

Hint: Assume for the sake of contradiction that the sequence is bounded above by some threshold $\lambda$. What can we say about the differences $x_{n+1}-x_n$? On one hand, $x_{n+1}-x_n = 1/x_n \geq 1/\lambda$, but...

Answer 3

The sequence is increasing indeed. Since $$x_{n+1} - x_n = \frac{1}{x_n} > 0$$ We can show that it is not bounded. Otherwise, it converges, say to $l$. Then we have $$ l = l + 1/l,$$ which is a contradiction.

Answer 4

Extending Clark's answer.

$x_{n+1}^2 =x_n^2+2+\dfrac1{x_n^2} $ so $x_{n+1}^2-x_n^2 \ge 2$.

Summing, $x_{n+1}^2-x_1^2 \ge 2n$ so $x_{n+1}^2 \ge x_1^2 + 2n = 2n+a^2 $ so $x_{n+1} \ge \sqrt{2n+a^2} \gt \sqrt{2n} $.

Therefore $x_{n+1} =x_n+\frac{1}{x_n} \le x_n+\frac{1}{\sqrt{2(n-1)}} $ or $x_{n+1}- x_n \le \frac{1}{\sqrt{2(n-1+a^2)}} \lt \frac{1}{\sqrt{2(n-1)}} $.

Summing,

$\begin{array}\\ x_{m+1}- x_2 &=\sum_{n=2}^m (x_{n+1}- x_n)\\ &\lt \sum_{n=2}^m\frac{1}{\sqrt{2(n-1)}}\\ &= \dfrac1{\sqrt{2}}\sum_{n=1}^{m-1}\frac{1}{\sqrt{n}}\\ &= \dfrac1{\sqrt{2}}(1+\sum_{n=2}^{m-1}\frac{1}{\sqrt{n}})\\ &< \dfrac1{\sqrt{2}}(1+\int_1^m \dfrac{dt}{t^{1/2}})\\ &= \dfrac1{\sqrt{2}}(1+\dfrac{t^{1/2}}{1/2}|_1^m)\\ &= \dfrac1{\sqrt{2}}(1+2(\sqrt{m}-1))\\ &= \dfrac1{\sqrt{2}}(2\sqrt{m}-1)\\ &= \sqrt{2m}-\dfrac1{\sqrt{2}}\\ \end{array} $

so that $\sqrt{2m} \lt x_{m+1} \lt \sqrt{2m}+x_2-\dfrac1{\sqrt{2}} $.

Extending this, I am sure that we can prove that $\lim_{m \to \infty} (x_m-\sqrt{2m}) $ exists (and I have a feeling that I already have), but I'll leave it at this.

Answer 5

I would like to suggest an another way, by considering that the sequence is monotone increasing : you can show this by seeing that

$$x_{n+1}=x_{n}+\frac{1}{x_{n}}>x_{n}$$

Now we have that $x_{1}<x_{2}<\cdots<x_{n}<x_{n+1}<\cdots$, consider that $$ \begin{align} &x_{n+1}=x_{n}+\frac{1}{x_{n}}=x_{n-1}+\frac{1}{x_{n-1}}+\frac{1}{x_{n}} \text{ continuing so gives} \\ &x_{n+1}=a+\sum_{i=2}^{n}\frac{1}{x_{i}} \end{align} $$ Assuming it's bounded then it must be convergent by the convergence theorem $\exists M\in \mathbb{R}$ such that $x_{n}<M$ and we know that $x_{1}<\cdots<x_{n}$ therefore we see that $\frac{1}{M}<\frac{1}{x_{i}}$ for $i=1,2,\cdots,n$ therefore we can do the following

$$x_{n+1}>\sum_{i=2}^{n}\frac{1}{M}+a=\frac{n-1}{M}+a$$ Therefore the $1$-tail is divergent which means the sequence $(x_{n})$ is divergent Thus a contradiction