Let $x_1=a>0$ and $x_{n+1}=x_{n}+\frac{1}{x_{n}};n>1$. Then does the sequence $(x_n)$ converge?

The sequence $(x_n)$ is increasing. But I could not show that it is bounded. Any hint in this regard will be appreciated.


Suppose it converge, and denote $\ell$ the limit. Clearly $\ell\neq 0$ (why ?)

Then, $$\lim_{n\to\infty }x_{n+1}=\lim_{n\to\infty }x_n+\lim_{n\to\infty }\frac{1}{x_n}\iff \ell=\ell+\frac{1}{\ell}\iff \frac{1}{\ell}=0\iff 1=0$$

contradiction !

  • $\begingroup$ I also thought like this...but can I show that it is unbounded? $\endgroup$ – Anupam Jan 13 '15 at 14:14
  • 1
    $\begingroup$ if you proved that the sequence is increasing and that the sequence has no limit, you proved that the sequence is unbounded ! $\endgroup$ – idm Jan 13 '15 at 14:25

Suppose $(x_n)$ is convergent $\iff$ Cauchy.
For each $n\in\mathbb{N}$ there exist a positive integer $K_n$ such that $$|x_p-x_q|\le\dfrac{1}{n}$$ whenever $p,q\ge K_n.$ For $p>q$ we have $$x_p-x_q=(x_p-x_{p-1})+(x_{p-1}-x_{P-2})+\cdots+(x_{q+1}-x_q)$$ $$x_p-x_q=\dfrac{1}{x_{p-1}}+\dfrac{1}{x_{p-2}}+\dots+\dfrac{1}{x_{q}}\le\dfrac{1}{n}$$ Hence $$\dfrac{1}{x_{r}}\le\dfrac{1}{n},\forall r\ge K_n.$$ This implies, for each $n\in\mathbb{N}$ there exist $K_n\in\mathbb{N}$ such that $x_r\ge n, \forall n\ge K_n.$
$(x_n)$ is unbounded. This is a contradiction.
Therefore $(x_n)$ is divergent.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.