text: # Dividing Divergent and Convergent Sequences
  
  Let $$a_n$$ be a diverging sequence and $$b_n$$ be a converging sequence. With that in mind does $$\frac{a_n}{b_n}$$ always diverge?
  
  If so, what is the proof for this?
  
  • One of the intermediate results in Real Analysis is that if $\langle r_n\rangle$ and $\langle s_n\rangle$ are two convergent sequences, then the sequence $$\langle {r_n} \times {s_n}\rangle$$ is also a convergent sequence. You can regard $a_n = b_n \times \frac{a_n}{b_n}$. Mar 24 '21 at 18:04
  
  Let $$b_n$$ converge to $$l_1$$, and suppose $$\frac{a_n}{b_n}$$ converges to $$l_2$$. Then by the Algebraic Limit Theorem we have$$\lim_{n\to \infty}b_n\cdot \frac{a_n}{b_n} = l_1l_2.$$ But $$b_n \cdot \frac{a_n}{b_n} =a_n,$$ which is divergent, thus giving us a contradiction. Therefore $$\frac{a_n}{b_n}$$ is divergent.
  • Thank you, following on from this if the opposite were to happen $a_n$ is convergent and $b_n$ is divergent will $\frac{a_n}{b_n}$ always either be divergent or converge to zero? Mar 24 '21 at 20:10
explanation: The document contains a math question about the convergence or divergence of the sequence \(\frac{a_n}{b_n}\) given that \(a_n\) is a diverging sequence and \(b_n\) is a converging sequence. It also includes an answer to this question, providing a proof by contradiction. Additionally, there is a follow-up question asking about the case where \(a_n\) is convergent and \(b_n\) is divergent.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
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  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # Dividing Divergent and Convergent Sequences
  
  Let $$a_n$$ be a diverging sequence and $$b_n$$ be a converging sequence. With that in mind does $$\frac{a_n}{b_n}$$ always diverge?
  
  If so, what is the proof for this?
  
  • One of the intermediate results in Real Analysis is that if $\langle r_n\rangle$ and $\langle s_n\rangle$ are two convergent sequences, then the sequence $$\langle {r_n} \times {s_n}\rangle$$ is also a convergent sequence. You can regard $a_n = b_n \times \frac{a_n}{b_n}$. Mar 24 '21 at 18:04
  
  Let $$b_n$$ converge to $$l_1$$, and suppose $$\frac{a_n}{b_n}$$ converges to $$l_2$$. Then by the Algebraic Limit Theorem we have$$\lim_{n\to \infty}b_n\cdot \frac{a_n}{b_n} = l_1l_2.$$ But $$b_n \cdot \frac{a_n}{b_n} =a_n,$$ which is divergent, thus giving us a contradiction. Therefore $$\frac{a_n}{b_n}$$ is divergent.
  • Thank you, following on from this if the opposite were to happen $a_n$ is convergent and $b_n$ is divergent will $\frac{a_n}{b_n}$ always either be divergent or converge to zero? Mar 24 '21 at 20:10
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math question about the convergence or divergence of the sequence \(\frac{a_n}{b_n}\) given that \(a_n\) is a diverging sequence and \(b_n\) is a converging sequence. It also includes an answer to this question, providing a proof by contradiction. Additionally, there is a follow-up question asking about the case where \(a_n\) is convergent and \(b_n\) is divergent.
  
  Label: question-and-answer