Math Problem Statement

Prove that every convergent sequence is bounded

Solution

To prove that every convergent sequence is bounded, we will use the definition of convergence and show that this property implies boundedness.

Definitions:

  • Convergent Sequence: A sequence {an}\{a_n\} is said to converge to a limit LL if, for every ϵ>0\epsilon > 0, there exists an integer NN such that for all nNn \geq N, we have: anL<ϵ.|a_n - L| < \epsilon. This means that as nn increases, the terms of the sequence get arbitrarily close to LL.

  • Bounded Sequence: A sequence {an}\{a_n\} is bounded if there exists a real number MM such that for all nn, the absolute value of ana_n is less than or equal to MM. In other words, there is some upper bound MM such that: anMfor alln.|a_n| \leq M \quad \text{for all} \, n.

Proof:

Let {an}\{a_n\} be a convergent sequence, and let it converge to a limit LL. By the definition of convergence, for every ϵ>0\epsilon > 0, there exists an integer NN such that for all nNn \geq N, we have: anL<ϵ.|a_n - L| < \epsilon.

Let's choose ϵ=1\epsilon = 1. This means that there exists an integer NN such that for all nNn \geq N, we have: anL<1.|a_n - L| < 1. This implies that: an<L+1for allnN.|a_n| < |L| + 1 \quad \text{for all} \, n \geq N.

Now, consider the first N1N-1 terms of the sequence {an}\{a_n\}, namely a1,a2,,aN1a_1, a_2, \dots, a_{N-1}. These terms are finite in number, and therefore, they are bounded. Let M1M_1 be an upper bound for these first N1N-1 terms, i.e., there exists a constant M1M_1 such that: anM1forn=1,2,,N1.|a_n| \leq M_1 \quad \text{for} \, n = 1, 2, \dots, N-1.

Next, let M2=L+1M_2 = |L| + 1, which bounds the terms ana_n for nNn \geq N, i.e.: anM2fornN.|a_n| \leq M_2 \quad \text{for} \, n \geq N.

Thus, the sequence {an}\{a_n\} is bounded by the maximum of M1M_1 and M2M_2. Let M=max(M1,M2)M = \max(M_1, M_2). Therefore, for all nn, we have: anM.|a_n| \leq M.

Hence, the sequence {an}\{a_n\} is bounded.

Conclusion:

We have shown that if {an}\{a_n\} is a convergent sequence, then there exists a real number MM such that anM|a_n| \leq M for all nn. Therefore, every convergent sequence is bounded.


Would you like more details on this proof or any clarifications?

Here are 5 related questions for further exploration:

  1. What happens if a sequence diverges? Is it always unbounded?
  2. How do we formally define the limit of a sequence?
  3. Can a bounded sequence still fail to converge?
  4. What is the difference between a convergent sequence and a Cauchy sequence?
  5. How do we use the ϵδ\epsilon-\delta definition in proving convergence?

Tip: Whenever dealing with limits and sequences, visualizing a sequence on a number line can often help in understanding how it behaves as nn increases.

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Math Problem Analysis

Mathematical Concepts

Sequences
Convergence
Boundedness

Formulas

|a_n - L| < epsilon
|a_n| <= M

Theorems

Definition of Convergence
Definition of Bounded Sequence

Suitable Grade Level

College-level