# scaling a sequence that tends to infinity, also tends to infinity proof

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?

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.

• 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? – Mathlearner Sep 16 at 14:54
• @Mathlearner How do you know that $kC>C$? What if $k=\frac12$? – 5xum Sep 16 at 15:05
• I just thought of this myself. Of course, it is not true. I will rethink my argument. – Mathlearner Sep 16 at 15:06
• @Mathlearner Hint: If you can get $a_n> \frac{C}{k}$, then $k\cdot a_n > k\cdot \frac{C}{k}$. – 5xum Sep 16 at 15:07
• @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 $\overline{C}=\frac{C}{k}$. – 5xum Sep 17 at 5:30