Additional Bounded Sequence Proofs

# Additional Bounded Sequence Proofs

Theorem 1: Let $(a_n)$ and $(b_n)$ be sequences. If $(a_n)$ converges to $0$ and $(b_n)$ is bounded then $(a_nb_n)$ converges to $0$. |

**Proof:**Let $\epsilon > 0$ be given.

- Since $(b_n)$ is bounded there exists an $M \in \mathbb{R}$, $M > 0$ such that $|b_n| \leq M$ for all $n \in \mathbb{N}$.

- Since $(a_n)$ converges to $0$ we have that for $\epsilon_1 = \frac{\epsilon}{M} > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then:

\begin{align} \quad |a_n| < \epsilon_1 = \frac{\epsilon_1}{M} \quad (*) \end{align}

- Then if $n \geq N$ we have that $(*)$ holds and so:

\begin{align} \quad |a_nb_n - 0| = |a_nb_n| = |a_n||b_n| < \epsilon_1 \cdot M = \frac{\epsilon}{M} \cdot M = \epsilon \end{align}

- Therefore $(a_nb_n)$ converges to $0$. $\blacksquare$