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The question asks:

"We know that the sequence \( \left( \frac{1}{n} \mid n \in \mathbb{N} \right) \) converges to 0. Is this sequence bounded?"

### Answer:

Yes, the sequence \( \left( \frac{1}{n} \right) \) is **bounded**.

Here's why:

1. A sequence is considered **bounded** if there exists a real number \( M \) such that for all terms \( a_n \) of the sequence, \( |a_n| \leq M \).
2. The sequence \( a_n = \frac{1}{n} \) for \( n \in \mathbb{N} \) is bounded above by 1, because:
   \[
   \frac{1}{n} \leq 1 \quad \text{for all } n \in \mathbb{N}.
   \]
3. Additionally, the sequence is bounded below by 0, because \( \frac{1}{n} \) is positive and approaches 0 as \( n \to \infty \).
4. Therefore, the sequence is bounded between 0 and 1, i.e., \( 0 \leq \frac{1}{n} \leq 1 \).

Thus, since the sequence is bounded above by 1 and bounded below by 0, it is **bounded**.

Would you like more details or further explanations?

Here are 5 related questions you might be interested in:
1. What does it mean for a sequence to converge?
2. How can we prove that a sequence is bounded mathematically?
3. What is the relationship between a sequence being bounded and converging?
4. Is every convergent sequence bounded?
5. Can a bounded sequence diverge?

**Tip:** A sequence that converges is always bounded, but not all bounded sequences necessarily converge.

We know that the sequence (1/n | n ∈ N) converges to 0. Is this sequence bounded?

Learn about the convergence and boundedness of the sequence a_n = 1/n, where n is a natural number. This problem discusses key concepts such as sequence convergence to 0 and its bounded nature. Suitable for Grades 10-12.

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