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 first show that the sequence is increasing. Observe that since \(n > 0\), then
$$
\begin{align*}
a_{n+1} - a_n = \frac{1}{(1+n)!} > 0
\end{align*}
$$
So \(a_{n+1} > a_n\) and hence, the sequence \(\{a_n\}\) is increasing. Moreover, recall that
$$
\begin{align*}
e = \sum_{n = 1}^{\infty} \frac{1}{n!} = 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots
\end{align*}
$$
Then clearly
$$
\begin{align*}
a_{n+1} - a_n = \frac{1}{(1+n)!} < e
\end{align*}
$$
for all \(n > 0\) so the sequence is bounded above. Since the sequence is monotone increasing and bounded above, it follows by the Monotone Convergence Theorem, that the sequence \(\{a_n\}\) is convergent as we wanted to show. \(\blacksquare\)
References
- Problem Statement Source: Aleph 0