I have a question in Analysis. Knowing that $x_{2n}$, $x_{2n-1}$, $x_{3n}$ converge, how can I show that $x_{n}$ converges?
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1Hint: You only need to know that two of those sequences converge, the other is a bit of a distraction. – Andrés E. Caicedo Nov 03 '13 at 01:33
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@AndresCaicedo I think we need the distraction too. – Lord Soth Nov 03 '13 at 01:40
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1@AndresCaicedo Yes e.g. $x_{2n}$ can converge to 1 and $x_{2n-1}$ can converge to -1. – meta_warrior Nov 03 '13 at 01:41
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1Ah, cute. Yes, no distractions. – Andrés E. Caicedo Nov 03 '13 at 02:08
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Hint: 1) $x_{2n}$ and $x_{3n}$ should converge to the same limit (by looking at a profitable subsequence of each). 2) A similar argument for another profitable choice of two of $x_{2n}, x_{2n-1}, x_{3n}$. 3) All three given subsequences actually converge to the same limit. We can now conclude via $x_{2n},x_{2n-1}$.
Lord Soth
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By using which subsequence can I show that all three given subsequences converge to the same limit? – Mary Star Nov 03 '13 at 12:09
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