Convergent sequence times a constant

Question

So I am planning on doing a self study for real analysis, but before that I want to get comfortable with sequence converging and how to prove them.

If a sequence $\{x_n\}$ is converging to $a$. How can we then tell that $\{kx_n\}$ converges to $ka$, where $k$ is just a constant? Also how can I better see what I need to choose $\epsilon$ or $N$ to satisfy the condition for any sequence?

I know that the construction of such a proof would look somewhat like this.

if $x_n\rightarrow a$, then given $\epsilon > 0$ $\exists n\ \in\mathbb{N}$ s.t if $n>N$ we have $|x_n-a|<\epsilon$. This is the part I usually faild to understand. What am I supposed to look for now? We stated that $\epsilon >0$ so now I need to find a suitable $N$, but how do I look for such a $N$? Similary the same would apply for $kx_n\rightarrow ka$ then $|kx_n-ka|<\epsilon$ Also also, I sometimes see $\frac{\epsilon}{2}$ Where does that come from?

Answer 1

For any $\epsilon>0$, you have $N$ from convergence of $(x_n)_n$.

Therefore if you take $\epsilon/|k|>0$ then you have corresponding $N$ such that $\forall n\geq N(|x_n-a|<\epsilon/|k|)$ hence $\forall n\geq N(|kx_n-ka|\leq\epsilon)$.

Answer 2

Here is what we are given from the convergence of $\{x_n\}$:

Let $\epsilon_1 >0 $ be arbitrary. Assume that $\{x_n\}$ converges to $a$. Thus it follows there exists $N \in \mathbb{N}$ such that if $n \geq N$ then \begin{equation*} |x_n - a| < \epsilon_1. \end{equation*}

Given this information, to show $\{kx_n\}$ converges to $ka$, we need to show that $|kx_n - ka| < \epsilon$, where $\epsilon$ is arbitrary. Now since we have control over $\epsilon_1$, we can fix $\epsilon_1 = \frac{\epsilon}{|k|}$. Thus it immediately follows that if $n \geq N$ (for the same $N \in \mathbb{N}$ as above) that \begin{equation*} |kx_n - ka| = |k||x_n-a| < |k|\epsilon_1 = |k|\frac{\epsilon}{|k|} = \epsilon. \end{equation*}

Also, echoing the comments on your question, we take $|k|$ instead of $k$ as we do not know if the constant is positive or negative.