Little-O Implies Big-O/Sequences

Theorem
Let $(a_n)$ and $(b_n)$ be sequences of real or complex numbers.

Let $a_n = o(b_n)$ where $o$ denotes little-O notation.

Then $a_n = O(b_n)$ where $O$ denotes big-O notation.

Proof
Because $a_n = o(b_n)$, there exists $n_0 \in\N$ such that $|a_n| \leq 1 \cdot |b_n|$ for $n\geq n_0$.

Thus $a_n = O(b_n)$.