If the sequence $s_n$ converges to $s$, then for all $k \in \mathbb{R}$ the sequence $ks_n$ converges to $ks$.
Proof: We assume that $k \neq 0$ since the result is trivial for $k = 0$ .
Let $\epsilon > 0$ and note that we have to proof $|ks_n - ks| < \epsilon$ for large n. Since $\lim s_n = s$, there exists $N \in \mathbb{N}$ such that
$n > N$ implies: $$|s_n - s| < \frac{\epsilon}{|k|}$$
which also implies: $$|ks_n - ks| < \epsilon$$
My question might sound dumb because the logic is pretty obvious because $\lim (ks_n) = k \cdot \lim s_n$. But I don't quite understand why does the proof starts with $|s_n - s| < \frac{\epsilon}{|k|}$? In another words, why do we have $\frac{\epsilon}{|k|}$?
Appreciate for the one who helps!
They did that because they want to end up with $|ks_n-ks|< \epsilon$, so they can show that $\lim\limits_{n\to \infty}ks_n = ks$.