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