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

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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.

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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...

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  • $\begingroup$ How do we assume this? $\endgroup$ – Unknown x Apr 29 '18 at 5:24
  • $\begingroup$ @ManeeshNarayanan For the sake of contradiction. $\endgroup$ – spaceisdarkgreen Apr 29 '18 at 5:27
  • $\begingroup$ How do we get contradiction here?From this, we can prove that sequence is increasing. and assume it is bounded. but it is not Cauchy. I think this might be the argument. am I right? $\endgroup$ – Unknown x Apr 29 '18 at 5:30
  • $\begingroup$ @ManeeshNarayanan Y.Hu's answer (which I see you have now accepted) shows how to apply this exact argument to give a contradiction. $\endgroup$ – Carl Schildkraut Apr 29 '18 at 5:33
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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.

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