Little-O Times Big-O is Little-O/Sequences

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

Let:
 * $a_n = \map \OO {b_n}$
 * $c_n = \map {\mathcal o} {d_n}$

where:
 * $\OO$ denotes big-$\OO$ notation
 * $\mathcal o$ denotes little-$\mathcal o$ notation.

Then:
 * $a_n c_n = \map {\mathcal o} {b_n d_n}$

Proof
Let $\epsilon \in \R_{> 0}$.

Since $a_n = \map \OO {b_n}$:
 * $\exists c \in \R: c \ge 0: \exists n_0 \in \N: \paren {n \ge n_0 \implies \size {a_n} \le c \cdot \size {b_n} }$

Since $c_n = \map {\mathcal o} {d_n}$:
 * $\exists n_1 \in \N: \paren {n \ge n_1 \implies \size {c_n} \le \dfrac \epsilon {c + 1} \cdot \size {d_n} }$

Thus for $n \ge \max \set {n_0, n_1}$:

Thus $a_n c_n = \map {\mathcal o} {b_n d_n}$.