prove - if a tends to L then Ma tends to ML Suppose that the sequence $(a_n)_{n=1}^{\infty}$ is such that $a_n \rightarrow L$ as $n\rightarrow\infty$. 

Prove that for any $M&gt;0, M \in \mathbb{R}$, we have that $Ma_n \rightarrow ML$ as $n \rightarrow \infty$.
Given $\epsilon&gt;0$ there exists $\overline{n}\in\mathbb{N}$ such that $|a_n-L|&lt;\frac{\epsilon}{M},\forall n\geq\overline{n}$, because by hypothesys $a_n$ converges to $L$. Now, for every $n\geq\overline{n}$, you'll get

$$|Ma_n -ML|=M|a_n-L|\leq M\frac{\epsilon}{M}=\epsilon$$

This proves $Ma_n\rightarrow ML$
Let $\varepsilon&gt;0$ be given. By the assumption, we can find an $N$ such that
$n\geqslant N \Rightarrow |a_{n} - L| &lt; \varepsilon /M$.
then,
$n \ge N  \Rightarrow |Ma_{n}−ML|=M|a_{n} − L|&lt; M*ϵ/M=ϵ$
which shows that if the sequence $\{a_{n}\}_{n=1}^\infty$ tends to limit $L$ as $n\to\infty$ then, for any fixed number $M&gt;0$, the sequence $\{Ma_{n}\}_{n=0}^\infty$ tends to the limit $ML$.





source: Course Introduction to Mathematical Thinking / Dr Keith Devlin