Math Problem Statement

Consider the Cauchy sequences (an)n∈N and (bn)n∈N, show that

  1. (an + bn)n∈N,

Solution

To show that (an+bn)nN(a_n + b_n)_{n \in \mathbb{N}} is a Cauchy sequence, we need to use the definition of a Cauchy sequence.

Definition of a Cauchy Sequence:

A sequence (xn)nN(x_n)_{n \in \mathbb{N}} is Cauchy if for every ϵ>0\epsilon > 0, there exists an integer NN such that for all m,nNm, n \geq N, xnxm<ϵ.|x_n - x_m| < \epsilon. Now, we are given that both (an)nN(a_n)_{n \in \mathbb{N}} and (bn)nN(b_n)_{n \in \mathbb{N}} are Cauchy sequences. This means:

  • For every ϵ>0\epsilon > 0, there exists an integer N1N_1 such that for all m,nN1m, n \geq N_1, anam<ϵ2.|a_n - a_m| < \frac{\epsilon}{2}.
  • Similarly, for every ϵ>0\epsilon > 0, there exists an integer N2N_2 such that for all m,nN2m, n \geq N_2, bnbm<ϵ2.|b_n - b_m| < \frac{\epsilon}{2}.

Now, let N=max(N1,N2)N = \max(N_1, N_2). For all m,nNm, n \geq N, we have: anam<ϵ2andbnbm<ϵ2.|a_n - a_m| < \frac{\epsilon}{2} \quad \text{and} \quad |b_n - b_m| < \frac{\epsilon}{2}.

Step-by-Step Proof:

Consider the sequence (an+bn)nN(a_n + b_n)_{n \in \mathbb{N}}. We need to show that this sequence is Cauchy. For m,nNm, n \geq N, observe the following: (an+bn)(am+bm)=(anam)+(bnbm).|(a_n + b_n) - (a_m + b_m)| = |(a_n - a_m) + (b_n - b_m)|.

By the triangle inequality, we have: (an+bn)(am+bm)anam+bnbm.|(a_n + b_n) - (a_m + b_m)| \leq |a_n - a_m| + |b_n - b_m|.

Now, using the fact that both anam<ϵ2|a_n - a_m| < \frac{\epsilon}{2} and bnbm<ϵ2|b_n - b_m| < \frac{\epsilon}{2} for all m,nNm, n \geq N, we get: (an+bn)(am+bm)ϵ2+ϵ2=ϵ.|(a_n + b_n) - (a_m + b_m)| \leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.

Therefore, for all m,nNm, n \geq N, we have: (an+bn)(am+bm)<ϵ.|(a_n + b_n) - (a_m + b_m)| < \epsilon.

This shows that the sequence (an+bn)nN(a_n + b_n)_{n \in \mathbb{N}} is a Cauchy sequence, since for every ϵ>0\epsilon > 0, there exists an integer NN such that for all m,nNm, n \geq N, the difference (an+bn)(am+bm)|(a_n + b_n) - (a_m + b_m)| is less than ϵ\epsilon.

Conclusion:

(an+bn)nNis a Cauchy sequence.(a_n + b_n)_{n \in \mathbb{N}} \quad \text{is a Cauchy sequence.}

Would you like a deeper explanation on any part of this proof or clarification? Here are five related questions:

  1. What properties of Cauchy sequences are important for proving convergence?
  2. Can the sum of two non-Cauchy sequences ever form a Cauchy sequence?
  3. How would this proof change if one sequence were not Cauchy?
  4. What role does the triangle inequality play in proofs involving Cauchy sequences?
  5. How does the concept of Cauchy sequences relate to completeness in metric spaces?

Tip: When adding two Cauchy sequences, the triangle inequality helps break down complex expressions into manageable parts.

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Math Problem Analysis

Mathematical Concepts

Cauchy Sequences
Metric Spaces
Convergence
Triangle Inequality

Formulas

|x_n - x_m| < ε
|(a_n + b_n) - (a_m + b_m)| ≤ |a_n - a_m| + |b_n - b_m|

Theorems

Cauchy Sequence Theorem
Triangle Inequality

Suitable Grade Level

Undergraduate Level - Real Analysis