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

Question

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)!

Answer 1

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 $\lim_{n\to +\infty}n x_n =\lim_{n\to +\infty}n c = ± \infty$

Answer 2

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.