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The sequenc \(\Large \{ \frac{1}{n} \}\)
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A) unbounded and convergent |
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B) bounded and convergent |
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C) bounded and divergent |
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D) unbounded land divergent |
Correct Answer:
| B) bounded and convergent |
Description for Correct answer:
\(x_{n}=\frac{1}{n}\forall n\in N,\) for all \(n\in N, 1\le x_{n}\) and \(n\in N,\ 0\le 0\)
\(\Rightarrow \{\frac{1}{n}\}\) is bounded.
\(x_{n}=\frac{1}{n}\) and \(x_{n+1}=\frac{1}{n+1}\)
\(\Rightarrow x_{n+1}-x_{n} \Large \frac{1}{n+1}-\frac{1}{n}=\frac{n-n-1}{(n+1)n}\)
\(\Large =\frac{-1}{n(n+1)}\)
\(\Rightarrow x_{n+1}-x_{n}\le 0\)
\(\Rightarrow x_{n+1}-x_{n}\) for all \(n\in N\), decreasing sequence.
Since is bounded and monotone sequence.
\(\therefore \{\frac{1}{n}\}\) is convergent.
\(x_{n}=\frac{1}{n}\forall n\in N,\) for all \(n\in N, 1\le x_{n}\) and \(n\in N,\ 0\le 0\)
\(\Rightarrow \{\frac{1}{n}\}\) is bounded.
\(x_{n}=\frac{1}{n}\) and \(x_{n+1}=\frac{1}{n+1}\)
\(\Rightarrow x_{n+1}-x_{n} \Large \frac{1}{n+1}-\frac{1}{n}=\frac{n-n-1}{(n+1)n}\)
\(\Large =\frac{-1}{n(n+1)}\)
\(\Rightarrow x_{n+1}-x_{n}\le 0\)
\(\Rightarrow x_{n+1}-x_{n}\) for all \(n\in N\), decreasing sequence.
Since is bounded and monotone sequence.
\(\therefore \{\frac{1}{n}\}\) is convergent.
Part of solved Real Analysis questions and answers : >> Elementary Mathematics >> Real Analysis
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