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

show that {a_n }:=(1/(1!))−(1/(2!))+(1/(3!))−(1/(4!))...+(((−1)^(n+1) )/(n!)) is a cauchy sequence. my attempt: let ε>0 we have ∣a_m −a_n ∣=∣(((−1)^(n+2) )/((n+1)!))+(((−1)^(n+3) )/((n+2)!))+...+(((−1)^(m+1) )/(m!))∣ ≤∣(((−1)^(n+2) )/((n+1)!))∣+∣(((−1)^(n+3) )/((n+2)!))∣+...+∣(((−1)^(m+1) )/(m!))∣ =(1/((n+1)!))+(1/((n+2)!))+...+(1/(m!))<(1/(n!))≤(1/n)<ε. (1/n)→0 as n→∞, so no matter any ε>0 (1/n)<ε eventually. as long as n^∗ >(1/ε) ∣a_m −a_n ∣<ε ∀n≥n^∗ .