Suppose that ${a_n}$ converges to a. prove the sequence ${ca_n}$ converges to ca, where c is any constant.

Here is the start of my proof:

By definition of convergence of a sequence, $|ca_n - ca| < \epsilon$ where $\epsilon > 0$ Then we can factor to get $c|a_n - a| < \epsilon$

I am stuck at this point, where do I go from here?


The trick is that since we know that $a_n$ converges to $a$, we may choose how close they come together, that is, we may make $|a_n-a|$ very small:

Suppose $a_n$ converges to $a$. What this means is that for every $\epsilon'>0$ there exists an $N'\in\mathbb{N}$ such that for all $n>N'$, $|a_n-a|<\epsilon'$.

Now let $\epsilon>0$, and choose $\epsilon'=\frac{\epsilon}{|c|}$. Now choose $N=N'$ as above. Then for any $n>N$ we have:

$|ca_n-ca|=|c||a_n-a|<|c|\frac{\epsilon}{|c|}=\epsilon$. Completing the proof.


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