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For $x_n \gt 0$ $\forall n=1,2,3,...$ Prove $$\lim\limits_{n\rightarrow +\infty}x_n=0$$ iff $$\lim\limits_{n\rightarrow +\infty}\frac{1}{x_n}=+\infty$$ Working proof:

$\forall M>0, \exists N$ such that $x_n>M$, $\forall n \ge N$

$x_n>M\Rightarrow(1/x_n)<(1/M)$

Choose $N=[M]+1$ $\Rightarrow$ $\frac{1}{x_n}<(1/x_N)=(1/x_{[M]+1})$

I have no idea what the next step would be or if I'm even headed in the right direction...

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    $\begingroup$ There is no other conditions like $x_n \gt 0$? $\endgroup$ – user261263 Dec 11 '17 at 18:18
  • $\begingroup$ $x_n>0$ $\forall n=1,2,3,...$ Sorry for that omission $\endgroup$ – Garrett Y. Dec 11 '17 at 18:20
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Suppose $x_n \to 0$. Let $M$ in $\mathbb R+$, there exists $N$ such that for all $n ≥ N, |x_n| ≤ 1/M,$ so for all $n ≥ N, |1/x_n| ≥ M$ (since $x > 0 \to 1/x$ decreases). So $1/x_n \to +\infty$

The same for the other direction.

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NOt true take $$x_n =\frac{(-1)^n}{n}$$

Now Edit:

Let $A>0$ since $x_n\to0$ there exists $N\in\Bbb N$ such that, $n\ge N$ one has $$0<x_n<\frac{1}{A}~~~\forall n>N$$ that is $$\frac{1}{x_n}>A~~~\forall n>N$$

Hence $\frac{1}{x_n}\to+\infty$

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  • $\begingroup$ @Lucas what ??? it depend on what the OP means. and which notation he use $\endgroup$ – Guy Fsone Dec 11 '17 at 18:08
  • $\begingroup$ For me $\to \infty$ means that $|x_n| \to + \infty$ $\endgroup$ – user371663 Dec 11 '17 at 18:09
  • $\begingroup$ @Lucas that is French notation as far as I know . and the OP did not use abs $\endgroup$ – Guy Fsone Dec 11 '17 at 18:10
  • $\begingroup$ @Lucas I disagree. He would say $|x_n|$ if he meant that. $\endgroup$ – Shashi Dec 11 '17 at 18:11
  • $\begingroup$ Indeed, the statement as written is untrue $\endgroup$ – user370967 Dec 11 '17 at 18:11

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