Problem #8

Question: prove (from the definition of a limit of a sequence) that if the sequence $\{a_n\}_{n=1}^\infty$ tends to limit $L$ as $n \rightarrow \infty$, then for any fixed number $M > 0$, the sequence $\{Ma_n\}_{n=1}^\infty$ tends to the limit $ML$.

Claim: if the sequence $\{a_n\}_{n=1}^\infty$ tends to limit $L$ as $n \rightarrow \infty$, then for any fixed number $M > 0$, the sequence $\{Ma_n\}_{n=1}^\infty$ tends to the limit $ML$.
Proof: let's write the formal definition of the claim. Any fixed number $M > 0$ given, also we know the fact that $\lim \limits_{x\to \infty}a_n = L$ and based on it we should proof that $$(\forall \epsilon > 0)(\exists n \in {N})(\forall m \geq n)[|Ma_m - ML| < \epsilon]$$ . By the definition of limit, $\lim \limits_{x\to \infty}a_n$ can be written as
$$(\forall \epsilon > 0)(\exists n \in {N})(\forall m \geq n)[|a_m - L| < \epsilon]$$
and, as it works for all $\epsilon > 0$, it works for $\frac {\epsilon}{M}$ and there is exists $n \in {N}$ such that
$$(\forall m \geq n)[|a_m - L| < \frac {\epsilon}{M}]$$
then, if we multiply both parts of inequality to $M$,
$$(\forall m \geq n)[M|a_m - L| < \epsilon]$$
and, by algebra,
$$(\forall m \geq n)[|Ma_m - ML| < \epsilon]$$
and it means that whenever an arbitrary $\epsilon > 0$ is given, there is exists $n \in {N}$ such that $(\forall m \geq n)[|Ma_m - ML| < \epsilon]$ and that prove the claim. $\blacksquare$