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I am trying to prove that if we have a sequence$(a_{n}) \rightarrow \infty$, then $(ka_{n}) \rightarrow \infty, k>0$ and $(ka_{n}) \rightarrow - \infty, k<0.$

Attempt:

Suppose $(a_{n}) \rightarrow \infty.$ Then for $C >0, \exists N_{1}$ such that $a_{n} > C$ for some $n > N_{1}$. If we now consider a fixed $k > 0$ then $a_{n} > \frac{C}{k}$ hence $ka_{n} > k \frac{C}{k} > C$. therefore it goes to infinity.

Similarly, suppose $(a_{n}) \rightarrow \infty$. Then for $C > 0, \exists N_{2}$ such that $a_{n} > C$ for some $n > N_{2}$. If we consider a fixed $k < 0$ then $a_{n} < \frac{C}{k} < k\frac{C}{k} < C$. therefore it goes to negative infinity.

EDIT:

I'm not sure if this does hold for $k = \frac{1}{2}$ say.

Do I perhaps need to consider fixed $k < 1$ as a separate case?

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Suppose $(a_{n}) \rightarrow \infty.$ Then for $C >0, \exists N_{1}$ such that $a_{n} > C$ for some $n > N_{1}$. If we now consider some $k > > 0$ then $ka_{n} > kC$ therefore the inequality still holds and we are done.

You cannot "consider some $k>>0$". You have a fixed value of $k$, and for that particular value of $k$ (that you cannot control), you need to prove that the sequence $(k\cdot a_n)$ converges. In other words, you need to prove the following statement:

$$\forall C>0 \exists N : \forall n>N: k\cdot a_n > C$$

You did not prove that as of yet.

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  • $\begingroup$ If I fix $k > 0$. Then i consider $ka_{n}$, can I not say that $ka_{n} > kC > C$ since in the case when $k > 0$ the inequality signs are preserved? $\endgroup$ – Mathlearner Sep 16 at 14:54
  • $\begingroup$ @Mathlearner How do you know that $kC>C$? What if $k=\frac12$? $\endgroup$ – 5xum Sep 16 at 15:05
  • $\begingroup$ I just thought of this myself. Of course, it is not true. I will rethink my argument. $\endgroup$ – Mathlearner Sep 16 at 15:06
  • $\begingroup$ @Mathlearner Hint: If you can get $a_n> \frac{C}{k}$, then $k\cdot a_n > k\cdot \frac{C}{k}$. $\endgroup$ – 5xum Sep 16 at 15:07
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    $\begingroup$ @Mathlearner Look ok to me. You can't just assume $a_n>\frac{C}{k}$, however, from the properties of $a_n$, you can prove that $a_n>\frac{C}{k}$ for all $n$ above a certain $N$. You know this because no matter what $\overline{C}$ you can find some $N$ such that $a_n>\overline{C}$ for $n>N$. In particular, this is also true for $\overline{C}=\frac{C}{k}$. $\endgroup$ – 5xum Sep 17 at 5:30

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