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Thread: proof for calc 5

  1. #1
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    proof for calc 5

    Suppose that {a_n} is a divergent sequence. Show that it has a subsequence {a_{n_k}} with 1+ n_k < n_{k+1} which also diverges

    can anyone give me a concise proof for this
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  2. #2
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    You could try an argument by contradiction. Suppose the result is false, so that every such subsequence converges. In particular, the subsequences a_{2n} and a_{2n+1} (of even-numbered and odd-numbered terms terms in the sequence) both converge, say a_{2n}\to l_1 and a_{2n+1}\to l_2. The subsequence a_{3n} also converges, and it's easy to see that its limit has to be both l_1 and l_2. So that means that l_1=l_2, and the whole sequence has to converge to that limit.
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