Proof verification of a convergent sequence

Question

If the sequence $<a_n>$ converges to $A$ and $k\in\mathbb{R}$, constant, then the sequence $<k a_n>$ converges to $kA$.

Attempted proof - Let $\epsilon > 0$ be given. Since $a_n\to A$ there exists an $N\in\mathbb{N}$ such that

$$|a_n - A| < \frac{\epsilon}{2|k|} \ \ \text{for} \ \ n > N$$

So,

\begin{align*} |ka_n - kA + kA - ka_n| &= |(a_n - A)k + (-(a_n - A)k)|\\ &\leq |a_n - A||k| + |-(a_n - A)||k|\\ &< \frac{\epsilon}{2|k|}|k| + \frac{\epsilon}{2|k|}|k|\\ &= \frac{\epsilon}{2} + \frac{\epsilon}{2}\\ &= \epsilon \end{align*}

I am not sure if this is completely right, any suggestions would help.

Answer 1

You showed that $|ka_n-kA+kA-ka_n|<\epsilon$, but in fact $|ka_n-kA+kA-ka_n|=0$,

so that inequality is not worth much.

In order to show that $<ka_n>$ converges to $kA,$

you want to show that for any $\epsilon>0$ there is an $N$ such that for all $n>N$, $|ka_n-kA|<\epsilon.$

You are given that $<a_n>$ converges to $A$,

so for any $\epsilon'>0$ there is an $N$ such that for all $n>N$, $|a_n-A|<\epsilon'.$

For $k\ne0,$ set $\epsilon'=\epsilon/|k|$ and you are done. You don't need to split $\epsilon$ in half.

As pointed out in a comment by @Theo Bendit, $k=0$ is a separate (but rather trivial) case.

Answer 2

This is close. To sharpen things up,you could take $N>0$ so that $\lvert a_n - A \rvert < \frac{\varepsilon}{\lvert k \rvert}$ for $n>N$ and consider $$\lvert ka_n - kA\rvert = \lvert k \rvert\lvert a_n-A\rvert <\varepsilon.$$