To prove that $\lim a_n = 0$, we will use the given properties and the Bolzano–Weierstrass theorem, which ensures the existence of a monotone subsequence for any bounded sequence. Here’s the step-by-step proof:

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Proof:

1. Assumptions and Setup:

• We are given that for all $n$, $a_n \neq 0$, which means $a_n \neq 0$ for any term of the sequence.
• The reciprocal sequence $1/a_n$ has no bounded subsequence.
• We aim to show that $\lim a_n = 0$.

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2. Contradiction Approach:

Assume, for contradiction, that $\lim a_n \neq 0$. This leads to two possible cases:

1. $(a_n)$ does not converge to zero but is bounded away from zero.
2. $(a_n)$ diverges (has terms going arbitrarily far from zero).

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3. Case 1: $(a_n)$ is bounded away from zero.

• If $(a_n)$ is bounded away from zero, there exists $\epsilon > 0$ such that $|a_n| \geq \epsilon$ for all $n$.
• This implies that $1/|a_n| \leq 1/\epsilon$ for all $n$, making $1/a_n$ a bounded sequence.
• By the Bolzano–Weierstrass theorem, $1/a_n$ has a bounded subsequence, which contradicts the given property that $1/a_n$ has no bounded subsequences.
• Hence, $(a_n)$ cannot be bounded away from zero.

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4. Case 2: $(a_n)$ diverges.

• If $(a_n)$ diverges, there exists a subsequence $(a_{n_k})$ such that $|a_{n_k}| \to \infty$.
• This implies that $1/|a_{n_k}| \to 0$, which is bounded (since $0$ is a bound).
• Again, this contradicts the property that $1/a_n$ has no bounded subsequences.

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5. Conclude $a_n \to 0$:

• Since $(a_n)$ cannot be bounded away from zero and cannot diverge, the only remaining possibility is that $(a_n)$ converges to $0$.
• Hence, $\lim a_n = 0$.

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Conclusion:

We have shown that $\lim a_n = 0$ using the given properties and the Bolzano–Weierstrass theorem.

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Do you have any questions or need clarification about this proof?

Expanding Questions:

1. Why does the Bolzano–Weierstrass theorem guarantee a monotone subsequence for a bounded sequence?
2. How does assuming $a_n$ is bounded away from zero lead to a contradiction?
3. Can you explain why $(a_n)$ diverging contradicts the unbounded property of $1/a_n$?
4. Could this result hold if $1/a_n$ has only one bounded subsequence instead of none?
5. How does this result generalize to sequences in $\mathbb{R}^n$?

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Tip:

To tackle proofs with sequences, always consider properties like boundedness, monotonicity, and convergence/divergence to break the problem into manageable cases.