Let {x n } and {y n } be Cauchy sequences and C a constance. Use following defin
ID: 2943167 • Letter: L
Question
Let {xn} and {yn} be Cauchy sequences and C a constance. Use following definition to show that {xn+ yn} is Cauchy, too.
Definition: A sequence {xn} is Cauchy if, for every epsilon > 0, there exists N such that:
|xn - xm| < epsilon whenever n > m > N.
Explanation / Answer
Let {x(n)} and {y(n)} be Cauchy sequences and C a constance. Use following definition to show that {x(n)+ y(n)} is Cauchy, too. Definition: A sequence {x(n)} is Cauchy if, for every epsilon > 0, there exists N such that: |x(n) - x(m)| m > N. Proof: By definition, we have for every epsilon1 > 0, there exists N1 such that: |x(n) - x(m)| N1. Also by definition, we have for every epsilon2 > 0, there exists N2 such that: |y(n) - y(m)| N2. So, for every epsilon* >0, we set epsilon1 = epsilon*/2, and epsilon2 = epsilon*/2. (Since epsilon1 and epsilon2 could be any epsilon greater than 0). Also we find N* such that N*=max(N1, N2). Based on this, we want to show that for every epsilon* > 0, there exists N* such that: |(x(n) +y(n)) - (x(m)+y(m))| N*, proving that {x(n)+y(n)} is Cauchy. Thus, we have |(x(n)+y(n))-(x(m)+y(m))| = |x(n)-x(m) + y(n) - y(m)|0, such that |[x(n)+y(n)]-[x(m)+y(m)]|N*. Thus, {x(n)+y(n)} is Cauchy. Hope this helps!Related Questions
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