# Tests for convergences

• Jul 8th 2009, 07:57 PM
Juancd08
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.
• Jul 12th 2009, 10:09 AM
mathemanyak
[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, 11:12 AM
HallsofIvy
[QUOTE=mathemanyak;338106]
Quote:

Originally Posted by Juancd08
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?