show-that-a-n-1-1-1-2-1-3-1-4-1-n-1-n-is-a-cauchy-sequence-my-attempt-let-gt-0-we-have-a-m-a-n-1-n-2-n-1-1-n-3-n-2-

Question Number 148635 by learner001 last updated on 29/Jul/21

show that {a_{n}} := \frac{1}{1!} - \frac{1}{2!} + \frac{1}{3!} - \frac{1}{4!} \dots + \frac{{(- 1)}^{n + 1}}{n!} is a cauchy

sequence .

my attempt :

let ϵ > 0 we have ∣ a_{m} - a_{n} ∣=∣ \frac{{(- 1)}^{n + 2}}{(n + 1)!} + \frac{{(- 1)}^{n + 3}}{(n + 2)!} + \dots + \frac{{(- 1)}^{m + 1}}{m!} ∣

⩽∣ \frac{{(- 1)}^{n + 2}}{(n + 1)!} ∣ + ∣ \frac{{(- 1)}^{n + 3}}{(n + 2)!} ∣ + \dots + ∣ \frac{{(- 1)}^{m + 1}}{m!} ∣

= \frac{1}{(n + 1)!} + \frac{1}{(n + 2)!} + \dots + \frac{1}{m!} < \frac{1}{n!} ⩽ \frac{1}{n} < ϵ .

\frac{1}{n} \to 0 as n \to \infty, so no matter any ϵ > 0 \frac{1}{n} < ϵ eventually .

as long as n^{*} > \frac{1}{ϵ} ∣ a_{m} - a_{n} ∣< ϵ \forall n ⩾ n^{*} .