Use the monotone convergence theorem to show that the following sequence is convergent: \(a_n = 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots + \frac{1}{n!}\)
Definitions and Theorems
Proof
We claim that \(\lim\limits_{n \rightarrow \infty} a_nb_n = 0\). To see this, we are given that \(\{b_n\}\) is bounded but we don’t know if \(b_n\) is convergent. Since it’s bounded, then we know there is an \(M \in \mathbb{R}\) such that
$$
\begin{align*}
\left| b_n \right| \leq M
\end{align*}
$$
We are also given that \(\lim\limits_{n \rightarrow \infty} a_n = 0\), then by definition, for all \(\epsilon > 0\), there exists some \(N \in \mathbb{N}\) such that when \(n \geq N\) then
$$
\begin{align*}
\left| a_n - 0 \right| &< \frac{\epsilon}{M} \\
\left| a_n \right| &< \frac{\epsilon}{M}
\end{align*}
$$
Then,
$$
\begin{align*}
\left| a_nb_n \right| &= \left| a_n \right| \left| b_n \right| \\
&< \frac{\epsilon}{M} \cdot M \\
&< \epsilon
\end{align*}
$$
Since \(\epsilon > 0\) is arbitrary, then we must have
$$
\begin{align*}
\lim\limits_{n \rightarrow \infty} a_nb_n = 0
\end{align*}
$$
as desired. \(\blacksquare\)
References