Convergence of Sequence If I know that the sequence $\{a_n\}$ converges to $a$,
then to prove that the sequence $\{ca_n\}$ (for a constant $c$) converges to $ca$, I would basically want $|ca_n - ca| < \epsilon$... So, to prove this we have 
$|c||a_n-a| < \epsilon$, so if we FIX $\epsilon$, we know since $a_n$ converges to $a$, there exists an $N$, such that whenever $n≥N$, then $|a_n-a| < \frac {\epsilon}{|c|}$. 
Thus we have $|ca_n-ca| < \epsilon$...
(sorry for lack of rigor in some parts)
So my question is: since we know that $a_n$ converges to $a$, we can make it small as we want. In this case we can make it smaller than $\frac{\epsilon}{|c|}$ using a specific $N$. But, is this specific $\epsilon$ still arbitrary? In order to prove convergence of a sequence, given any $\epsilon >0$, there exists an $N$ such that...
So how does this work for this proof then, since we are choosing a specific $\epsilon$? (but is it still arbitrary?)
 A: You just have to be careful in your own mind about when you are stating a definition versus when you are using a fact about a sequence that follows from that definition. Let's take it from the top:


*

*You want to show that for any $\epsilon$ there exists $N$ such that whenever $n\ge N$ then $|ca_n - ca| = |c|\cdot |a_n-a| < \epsilon$. So far so good.

*As you point out, this is the same as saying that for any $\epsilon$ there exists $N$ such that whenever $n\ge N$ then $|a_n-a| < \frac{\epsilon}{|c|}$.

*So how will you prove that statement? We want to prove that something holds for any $\epsilon>0$. So choose an arbitrary $\epsilon>0$; we want to find an $N$ that works for that $\epsilon$.

*You know that $a_n\to a$, so that means that for any $\epsilon'>0$ there is an $N$ such that whenever $n\ge N$ then $|a_n-a| < \epsilon'$. Since this statement holds for any $\epsilon'>0$, in particular it holds when $\epsilon' = \frac{\epsilon}{|c|}$.

*We conclude that there is some $N$ such that whenever $n\ge N$ then
$$|a_n-a| < \epsilon' = \frac{\epsilon}{|c|},$$
which is what you were trying to prove.

A: There is no need for this : Simply note that
$|ca_n-ca|=|c||a_n-a|<c\epsilon $
Try to understand the  inner meaning of convergence .It says the distance between the terms of the sequence get very small after a certain period .
Since $\epsilon $ is arbitrary and chosen very small so terms of the sequence $ca_n$ also get very close after a certain period (i.e. same as $a_n$) .
For technical purpose note that as $\epsilon $ is arbitrary so is $c\epsilon$ for any $c\in \mathbb R$
