Linear Combination of Null Sequences

Be $(a_n) \subset \mathbb{Q}$ and $(b_n) \subset \mathbb{Q}$ null sequences. We want to prove that the sequence $(c_n) = \lambda (a_n) + \mu (b_n$) is a null sequence, for any $\lambda, \mu \in \mathbb{Q}$.

As we know, as $(a_n)$ is a null sequence, then exists an $\epsilon \in \mathbb{Q}$ such that for all $\epsilon > 0$, exists an $N_a \in \mathbb{N}$ such that for all $n \geq N_a$, we have $|a_n| < \epsilon$. The same is valid for $(b_n)$, so exists an $\epsilon \in \mathbb{Q}$ such that for all $\epsilon > 0$, exists an $N_b \in \mathbb{N}$ such that for all $n \geq N_b$, we have $|b_n| < \epsilon$.

Taking $\max(N_a, N_b) = N$, we have that for all $n \geq N$, we have $|a_n| < \epsilon$ and $|b_n| < \epsilon$, so:

• For $\frac{\epsilon}{2|\lambda|} > 0$, exists an $N_a \in \mathbb{N}$ such that for all $n \geq N$, we have $|a_n| < \frac{\epsilon}{2|\lambda|}$.
• For $\frac{\epsilon}{2|\mu|} > 0$, exists an $N_b \in \mathbb{N}$ such that for all $n \geq N$, we have $|b_n| < \frac{\epsilon}{2|\mu|}$.

So, we can write:

Therefore, we have proved that the linear combination of null sequences is a null sequence.