How to show the $\sum1/k!$ is a Cauchy sequence?

Question

How do you show that the sequence $\displaystyle a_n =\sum_{k=0}^n\frac 1{k!}$ is a Cauchy sequence?

I know we can use $k! \ge 2^{k-1}$ for all $k>1$, $\displaystyle \sum_{k=m}^\infty \frac 1{2^k} \le \frac 1{2^{m-1}}$, but how to use this hint?

Answer 1

Notice that the sequence is defined as a partial sum. Letting $m,n \in \mathbb{N}$ with $m>n$, we have that $$a_m - a_n = \sum_{k=n+1}^m \frac{1}{k!}$$

This is where the hint comes in. Because $k! \geq 2^{k-1}$ this implies $$a_m - a_n = \sum_{k=n+1}^m \frac{1}{k!} \leq \sum_{k=n}^{m-1} \frac{1}{2^k} < \sum_{k=n}^\infty \frac{1}{2^k}=\frac{1}{2^{n-1}}$$

Notice that we have obtained a bound on $a_m - a_n$ for arbitrary indexes $m>n$ which depends only on the smaller index. This is enought to prove it is Cauchy.

Proof: Let $\varepsilon>0$ be given. Choose $N\in\mathbb{N}$ such that $\frac{1}{2^N}<\varepsilon$. For any $n,m\geq N$ we then have $$|a_n - a_m| = |a_n-a_N+a_N-a_m| \leq |a_n - a_N| + |a_m - a_N|$$ Since $n \geq N$ and $m \geq N$ using the property above both $a_n - a_N < \frac{1}{2^{N-1}}$ and $a_m - a_N < \frac{1}{2^{N-1}}$. Hence we can show $$|a_n - a_N| + |a_m - a_N| < \frac{1}{2^{N-1}} + \frac{1}{2^{N-1}} = \frac{1}{2^N} < \varepsilon$$

Answer 2

Proposition

Every convergent sequence is Cauchy.

Proof

Let $(X,d_{X})$ be a metric space and let $x_{n}\in X$ be a sequence which converges to $x_{0}$. Thus, for every $\varepsilon > 0$, there corresponds a natural number $n_{\varepsilon}\in\mathbb{N}$ such that \begin{align*} n\geq n_{\varepsilon} \Rightarrow d_{X}(x_{n},x_{0}) \leq \varepsilon/2 \end{align*}

Consequently, if we take $n\geq n_{\varepsilon}$ and $m\geq n_{\varepsilon}$, it results that \begin{align*} n,m\geq n_{\varepsilon} \Rightarrow d_{X}(x_{n},x_{m}) \leq d_{X}(x_{n},x_{0}) + d_{X}(x_{m},x_{0}) \leq \varepsilon/2 + \varepsilon/2 = \varepsilon \end{align*} whence we conclude that $x_{n}$ is Cauchy.

Proposition

The exponential function, defined by \begin{align*} \exp(x) := \sum_{n=0}^{\infty}\frac{x^{n}}{n!} \end{align*} converges for every real number $x\in\mathbb{R}$.

Proof

Let us apply the ratio test. One has that \begin{align*} \limsup_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right| = \limsup_{n\to\infty}\left|\frac{x^{n+1}}{(n+1)!}\times\frac{n!}{x^{n}}\right| = \limsup_{n\to\infty}\frac{|x|}{n+1} = 0 < 1 \end{align*} for every real number $x\in\mathbb{R}$.

Solution

At your case, it suffices to notice that \begin{align*} \lim_{n\to\infty}\sum_{k=0}^{n}\frac{1}{k!} = \exp(1) = e \end{align*} Hence $a_{n}$ is convergent, whence we conclude that it is Cauchy based on the previous result.

Hopefully this helps.