prove that this is a cauchy sequence

Question

Prove that $\{x_n\}$ = $e^{-n}$ is a Cauchy sequence.

I tried to prove this by proving that,

For all $ϵ>0$, there is a positive $N$ s.t. for all $n>N$, $|e^{-n}|< ϵ$
For all $ϵ>0$, there is a positive $N$ s.t. for all $m>N$, $|e^{-m}|< ϵ$

I want some help with combining above two results to form
For all $ϵ>0$, there is a positive $N$ s.t. for all $n,m>N$, $|e^{-n} - e^{-m} |< ϵ$

Answer 1

It converges to $0$! So it is in particular a Cauchy sequence (every convergent sequence is Cauchy).

Answer 2

Hint: If $m \ge n$, then $ 0 < e^{-m} \le e^{-n}$, and so $0 \le e^{-n}-e^{-m} < e^{-n}$.

Similarly, if $n \ge m$, then $0 < e^{-n} \le e^{-m}$, and so $-e^{-m} < e^{-n}-e^{-m} \le 0$.

Answer 3

If you insist on wanting to show that your sequence is Cauchy without using convergence:

Suppose $m \geq n \geq N$. In that case

$$|e^{-m} - e^{-n}| = e^{-n} - e^{-m} = e^{-n} (1 - e^{n - m}) \leq e^{-n} \leq e^{-N}.$$

Now for $\varepsilon > 0$, you can pick $N$ such that $e^{-N} \leq \varepsilon$.