prove-that-a-n-n-1-defined-by-a-n-1-2-1-6-1-n-n-1-is-cauchy-sequence-

Question Number 148609 by learner001 last updated on 29/Jul/21

prove that {(a_{n})}_{n ⩾ 1} defined by a_{n} = \frac{1}{2} + \frac{1}{6} + \dots + \frac{1}{n (n + 1)} is

cauchy sequence .

Commented by learner001 last updated on 29/Jul/21

This is what i tried .

let ϵ > 0 be given i need an n^{*} such that p \in N if \forall n ⩾ n^{*} then

∣ a_{n + p} - a_{n} ∣< ϵ .

∣ a_{n + p} - a_{n} ∣=∣ (\frac{1}{n + 1} - \frac{1}{n + 2}) + (\frac{1}{n + 2} - \frac{1}{n + 3}) + \dots + (\frac{1}{n + p} - \frac{1}{n + p + 1}) ∣

=∣ \frac{1}{n + 1} - \frac{1}{n + p + 1} ∣⩽∣ \frac{1}{n + 1} ∣ + ∣ \frac{1}{n + p + 1} ∣< \frac{1}{n} + \frac{1}{n + p} < \frac{1}{n} < ϵ

if n^{*} ⩾ \frac{1}{ϵ} then ∣ a_{n + p} - a_{n} ∣< ϵ \forall n ⩾ n^{*} .

Commented by learner001 last updated on 29/Jul/21

is this correct?