Prove or disprove: If the sequence $(x_{n})_{n\in\mathbb{N}}\subset \mathbb{R}$ is convergent then $(nx_{n})_{n\in\mathbb{N}}$ is divergent.

The statement is true.

(It would work for some exceptions, like when the limit of the sequence $(x_{n}$) was $0$. Then the sequence would be convergent, too.)

But we don't know what $n$ is, we know it's a natural number. It could be $+\infty$ and this multiplied with a value larger than $0$ equals $\infty $ which makes the second sequences $(nx_{n})$ divergent.

Is everything correct?

This task was taken from an old exam, you get 2 points for solving it correctly. How many points would you give me for this solution?

Edit: The statement is wrong because a counter-example is enough to disprove it (see the counter-example at the beginning in the brackets)!

  • $\begingroup$ The statement is true if $x_n$ is convergent to $l \neq 0$. $\endgroup$ – Dark Jul 28 '16 at 12:28
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    $\begingroup$ Some people say statement is true because they assume first sequence converges to zero. But what if it doesn't converge to zero...? We need to respect all cases, I think. But I'm very confused now. $\endgroup$ – cnmesr Jul 28 '16 at 12:29
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    $\begingroup$ The claim is that, regardless of the convergent sequence $\langle x_n\rangle$ chosen, the sequence $\langle nx_n\rangle$ is divergent. The claim is false, and a single counterexample is all that is needed, such as the constant $0$ sequence. $\endgroup$ – Cameron Buie Jul 28 '16 at 12:40
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    $\begingroup$ If it had been specified that the sequence converged to a non-zero limit, then the statement would be true. $\endgroup$ – Cameron Buie Jul 28 '16 at 12:41
  • $\begingroup$ Alright now I understand everything, thank you Cameron! $\endgroup$ – cnmesr Jul 28 '16 at 12:43

The statement is not true. If $x_n = 0$ which is convergent then $n x_n = 0$ which is also convergent.

For the non-zero case it is true. If $\lim_{n\to +\infty}x_n =c$, then $\lim_{n\to +\infty}n x_n =\lim_{n\to +\infty}n c = ± \infty$

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    $\begingroup$ What if $c<0$? It is better to say that $(c x_n)_{n \in \Bbb N}$ diverges, rather than say that it tends to $\infty$. $\endgroup$ – Alex M. Jul 28 '16 at 12:34
  • $\begingroup$ @AlexM. Sorry a typo. $\endgroup$ – Zack Ni Jul 28 '16 at 12:37
  • $\begingroup$ So that would mean that the complete statement is true, right? Or is the more correct answer "not necessarily"? $\endgroup$ – cnmesr Jul 28 '16 at 12:38
  • $\begingroup$ @cnmesr Yes with non-zero limit it is a sufficient condition but without it, it will be a neither s nor n condition. $\endgroup$ – Zack Ni Jul 28 '16 at 12:41
  • $\begingroup$ @cnmesr you are welcome and this statement should be in the problem rather than my answer ^-^. $\endgroup$ – Zack Ni Jul 28 '16 at 12:46

Suppose wlog that as $n \to \infty $, $x_n \to L > 0$. Then there exists some $N$ so that $n \geq N $ implies that $x_n> L/2$.

Then $n x_n > n (\frac {L}{2}) $, which diverges.


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