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Math Help - Tests for convergences

  1. #1
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    Tests for convergences

    SUM(arctan(a_n) converges if and only if SUM(a_n) converges.

    a_n in this case is a sequence of positive numbers.
    Last edited by Juancd08; July 9th 2009 at 07:39 PM.
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  2. #2
    Junior Member mathemanyak's Avatar
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    [quote=Juancd08;337083]SUM(arctan(a_n) converges if and only if SUM(a_n) converges.

    preposition
    1.Let arctan(a_n)=b_n => tan(b_n)=a_n
    2.ıf any sequence is converges than its subsequences also converge.
    Solution
    tan(b_n) is a subsequence of b_n. By the comparison princeple if b_n is converge then tan b_n also converge. Conversly now show that if tan(b_n) is converge than b_n converges. Similiarly arctan(a_n) is subsequence of a_n then right handed also satisfies. then proof is done
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  3. #3
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    [QUOTE=mathemanyak;338106]
    Quote Originally Posted by Juancd08 View Post
    SUM(arctan(a_n) converges if and only if SUM(a_n) converges.

    preposition
    1.Let arctan(a_n)=b_n => tan(b_n)=a_n
    2.ıf any sequence is converges than its subsequences also converge.
    Solution
    tan(b_n) is a subsequence of b_n. By the comparison princeple if b_n is converge then tan b_n also converge. Conversly now show that if tan(b_n) is converge than b_n converges. Similiarly arctan(a_n) is subsequence of a_n then right handed also satisfies. then proof is done
    How is tan(b_n) a subsequence of b_n? You are saying that, for any positive integer m, there exist a positive integer n such that b_n= tan(b_m)? I don't see that. Or are you simply asserting that tan(b_n) is dominated by bn: that tan(b_n)< b_n for all n?
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