a-n-n-n-1-let-gt-0-be-given-a-m-a-n-m-m-1-n-n-1-m-n-m-1-n-1-m-n-m-1-n-1-provided-m-gt-n-m-n-m-1-n-1-lt-m-1-m-1-n-1-1-n-1-lt-if-N-gt

Question Number 148519 by learner001 last updated on 28/Jul/21

a_{n} = \frac{n}{n + 1}

let ϵ > 0 be given, ∣ a_{m} - a_{n} ∣=∣ \frac{m}{m + 1} - \frac{n}{n + 1} ∣=∣ \frac{m - n}{(m + 1) (n + 1)} ∣= \frac{m - n}{(m + 1) (n + 1)} provided

m > n, \frac{m - n}{(m + 1) (n + 1)} < \frac{m + 1}{(m + 1) (n + 1)} = \frac{1}{n + 1} < ϵ . if N > \frac{1 - ϵ}{ϵ} then ∣ a_{m} - a_{n} ∣< ϵ \forall n, m ⩾ N

Commented by learner001 last updated on 28/Jul/21

can someone check if the proof I made is correct? I was trying to show that the sequence is Cauchy.