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

a_n in this case is a sequence of positive numbers.

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- Jul 8th 2009, 08:57 PMJuancd08Tests 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. - Jul 12th 2009, 11:09 AMmathemanyak
[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 - Jul 12th 2009, 12:12 PMHallsofIvy
[QUOTE=mathemanyak;338106]

**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?