Math Problem Statement

using the fact that the sum of two cauchy sequences is cauchy, prove that the sum of two convergent series a_k and b_k converges

Solution

To prove that the sum of two convergent series aka_k and bkb_k converges, we will use the fact that the sum of two Cauchy sequences is also a Cauchy sequence.

Step-by-Step Proof:

  1. Recall the definition of convergence for series:

    A series ak\sum a_k converges if the sequence of partial sums SN=k=1NakS_N = \sum_{k=1}^N a_k converges to a limit as NN \to \infty. Similarly, for the series bk\sum b_k, the partial sums TN=k=1NbkT_N = \sum_{k=1}^N b_k converge to a limit.

  2. Convergence implies Cauchy sequence:

    If a series ak\sum a_k converges, then the sequence of its partial sums SNS_N is a Cauchy sequence. That is, for any ϵ>0\epsilon > 0, there exists an integer N1N_1 such that for all n,m>N1n, m > N_1, SnSm<ϵ.|S_n - S_m| < \epsilon. Similarly, since bk\sum b_k converges, the sequence of partial sums TNT_N is also a Cauchy sequence. That is, for any ϵ>0\epsilon > 0, there exists an integer N2N_2 such that for all n,m>N2n, m > N_2, TnTm<ϵ.|T_n - T_m| < \epsilon.

  3. Sum of two convergent series:

    Consider the series (ak+bk)\sum (a_k + b_k), where we define the partial sums of the sum as UN=k=1N(ak+bk)U_N = \sum_{k=1}^N (a_k + b_k). We can write this as: UN=SN+TN,U_N = S_N + T_N, where SNS_N is the partial sum of ak\sum a_k, and TNT_N is the partial sum of bk\sum b_k.

  4. Show that UNU_N is a Cauchy sequence:

    To show that (ak+bk)\sum (a_k + b_k) converges, we need to prove that the sequence UNU_N is Cauchy. Let n,m>max(N1,N2)n, m > \max(N_1, N_2). Then: UnUm=(Sn+Tn)(Sm+Tm)=(SnSm)+(TnTm).|U_n - U_m| = |(S_n + T_n) - (S_m + T_m)| = |(S_n - S_m) + (T_n - T_m)|. Using the triangle inequality: UnUmSnSm+TnTm.|U_n - U_m| \leq |S_n - S_m| + |T_n - T_m|. Since SnSm<ϵ|S_n - S_m| < \epsilon and TnTm<ϵ|T_n - T_m| < \epsilon for n,m>N1,N2n, m > N_1, N_2, we have: UnUm<ϵ+ϵ=2ϵ.|U_n - U_m| < \epsilon + \epsilon = 2\epsilon. This shows that UNU_N is a Cauchy sequence.

  5. Conclusion:

    Since the sequence of partial sums UNU_N is Cauchy, and since every Cauchy sequence in R\mathbb{R} converges, it follows that the series (ak+bk)\sum (a_k + b_k) converges.

Thus, the sum of two convergent series ak\sum a_k and bk\sum b_k converges.


Would you like to go over the steps in more detail? Let me know if you have any questions.

Related Questions:

  1. What does it mean for a series to converge absolutely?
  2. How can you prove that a sequence is Cauchy?
  3. What are the implications of the Cauchy Criterion for series?
  4. How do we prove that the difference of two convergent series also converges?
  5. What is the relationship between Cauchy sequences and limits?

Tip: When working with series, remember that convergence implies the existence of a limit, and that partial sums play a crucial role in proving convergence.

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Math Problem Analysis

Mathematical Concepts

Cauchy Sequences
Convergence of Series
Partial Sums
Real Analysis

Formulas

|S_n - S_m| < ε
|T_n - T_m| < ε
U_N = S_N + T_N

Theorems

Cauchy Criterion for Series
Convergence of Series implies Cauchy Sequences

Suitable Grade Level

College-level (Advanced Undergraduate or Graduate)