Show that if (an)n∈N,(bn)n∈N are equivalent sequences, then (an)n∈N converges
if and only if (bn)n∈N converges

To prove that if two sequences \((a_n)_{n \in \mathbb{N}}\) and \((b_n)_{n \in \mathbb{N}}\) are **equivalent**, then \((a_n)_{n \in \mathbb{N}}\) converges if and only if \((b_n)_{n \in \mathbb{N}}\) converges, we need to follow these steps:

### Definition of Equivalent Sequences:
Two sequences \((a_n)_{n \in \mathbb{N}}\) and \((b_n)_{n \in \mathbb{N}}\) are said to be **equivalent** if:
\[
\lim_{n \to \infty} (a_n - b_n) = 0.
\]
This means the difference between the terms of the two sequences tends to zero as \(n\) approaches infinity.

### Proof:

#### (1) Suppose \((a_n)_{n \in \mathbb{N}}\) converges:

- Let \(\lim_{n \to \infty} a_n = L\) for some real number \(L\).
- Since \((a_n)\) and \((b_n)\) are equivalent, we have:
  \[
  \lim_{n \to \infty} (a_n - b_n) = 0.
  \]
  This implies that for any \(\epsilon > 0\), there exists an \(N_1 \in \mathbb{N}\) such that for all \(n \geq N_1\),
  \[
  |a_n - b_n| < \frac{\epsilon}{2}.
  \]
- Since \((a_n)\) converges to \(L\), there exists an \(N_2 \in \mathbb{N}\) such that for all \(n \geq N_2\),
  \[
  |a_n - L| < \frac{\epsilon}{2}.
  \]
- Now, for \(n \geq \max(N_1, N_2)\), using the triangle inequality:
  \[
  |b_n - L| \leq |b_n - a_n| + |a_n - L| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.
  \]
Thus, \(\lim_{n \to \infty} b_n = L\), showing that \((b_n)\) converges to the same limit \(L\).

#### (2) Suppose \((b_n)_{n \in \mathbb{N}}\) converges:

- Let \(\lim_{n \to \infty} b_n = M\) for some real number \(M\).
- Since \((a_n)\) and \((b_n)\) are equivalent, we have:
  \[
  \lim_{n \to \infty} (a_n - b_n) = 0.
  \]
  Thus, for any \(\epsilon > 0\), there exists an \(N_3 \in \mathbb{N}\) such that for all \(n \geq N_3\),
  \[
  |a_n - b_n| < \frac{\epsilon}{2}.
  \]
- Since \((b_n)\) converges to \(M\), there exists an \(N_4 \in \mathbb{N}\) such that for all \(n \geq N_4\),
  \[
  |b_n - M| < \frac{\epsilon}{2}.
  \]
- Now, for \(n \geq \max(N_3, N_4)\), using the triangle inequality:
  \[
  |a_n - M| \leq |a_n - b_n| + |b_n - M| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.
  \]
Thus, \(\lim_{n \to \infty} a_n = M\), showing that \((a_n)\) converges to the same limit \(M\).

### Conclusion:
We have shown that if \((a_n)\) converges, then \((b_n)\) converges to the same limit, and vice versa. Therefore, if two sequences \((a_n)\) and \((b_n)\) are equivalent, then \((a_n)\) converges if and only if \((b_n)\) converges.

Would you like a deeper explanation on any part of this proof? 

Here are some related questions you might find interesting:

1. What are Cauchy sequences, and how are they related to convergent sequences?
2. Can a divergent sequence be equivalent to a convergent sequence?
3. How does the definition of equivalence change if we consider complex sequences?
4. Is the property of equivalence of sequences transitive?
5. Can you provide examples of two equivalent sequences?

**Tip**: The triangle inequality is a powerful tool in proving convergence, especially when comparing two sequences.

Show that if (a_n)_(n∈N), (b_n)_(n∈N) are equivalent sequences, then (a_n)_(n∈N) converges if and only if (b_n)_(n∈N) converges.

Understand how to prove that two equivalent sequences converge if and only if one of them converges. This guide walks through the concept of equivalent sequences, limits, and the triangle inequality theorem, suitable for undergraduate calculus students.

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