Little-O Implies Big-O/Sequences

Theorem
Let $\sequence {a_n}$ and $\sequence {b_n}$ be sequences of real or complex numbers.

Let $a_n = \map o {b_n}$ where $o$ denotes little-O notation.

Then $a_n = \map O {b_n}$ where $O$ denotes big-O notation.

Proof
Because $a_n = \map o {b_n}$, there exists $n_0 \in \N$ such that $\size {a_n} \le 1 \cdot \size {b_n}$ for $n \ge n_0$.

Thus $a_n = \map O {b_n}$.