Question

Show that {xn} is Cauchy if lim yn = 0, and |xm −xk | ≤ yk for all k ∈ n and for all m ≥ k.

a) {1/n}
b) {1/2^n}
c) {1/sqrt(n)}
d) {1/log(n)}

Answer 1

To show that the sequence {xn} is Cauchy, we need to prove that for any positive value ε, there exists a positive integer N such that |xm - xn| < ε for all m, n ≥ N. By analyzing each sequence, we can conclude that the sequences {1/n}, {1/2^n}, {1/√n}, and {1/log(n)} are all Cauchy.

Step-by-step explanation:

To show that the sequence {xn} is Cauchy, we need to prove that for any positive value ε, there exists a positive integer N such that |xm - xn| < ε for all m, n ≥ N.

Let's consider each sequence one by one:

a) For the sequence {1/n}, we have |1/m - 1/n| = |(n-m)/(mn)| ≤ |n-m|/n^2 ≤ 1/n^2 < ε for N = 1/√ε.

b) For the sequence {1/2^n}, we have |1/2^m - 1/2^n| = 1/2^m - 1/2^n = 1/2^n(2^(n-m) - 1) ≤ 1/2^n < ε for N = log2(1/ε).

c) For the sequence {1/√n}, we have |1/√m - 1/√n| = |√n - √m|/√(mn) = |n-m|/√(mn) ≤ |n-m|/√n^2 = |n-m|/n < ε for N = 1/ε^2.

d) For the sequence {1/log(n)}, we have |1/log(m) - 1/log(n)| = |log(n) - log(m)|/log(m) = log(n/m)/log(m) ≤ log(n/m)/log(min(n,m)) ≤ 1/(?log(min(n,m))^2) ≤ 1/?log(N)^2 < ε for N = ceil(1/√ε).