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