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Series convergence "by Parts
Supose:
[tex] \sum c_n = \sum (a_n+b_n) [/tex] (*1)
[tex] \sum a_n [/tex] is conditionaly convergent (*2)
[tex] \sum b_n [/tex] is absolutly convergent (*3)
And I have seen this proof: [Proving [tex] \sum c_n[/tex] is conditionally convergent]
From (*1) and (*2) [tex]\Rightarrow[/tex] [tex] \sum c_n[/tex] its convergent [this one I understand, basic properties of series]
But now they do something like this: [proof by contradiction]
Supose [tex] \sum |c_n| [/tex] is convergent
so [tex] |a_n|\leq|c_n-b_n|\leq|c_n|+|b_n| [/tex] (How they "jump" to this conclusion?!)
and now the use comparison test to show that [tex]\sum c_n[/tex] is conditionally convergent. [No problems from here]
Supose:
[tex] \sum c_n = \sum (a_n+b_n) [/tex] (*1)
[tex] \sum a_n [/tex] is conditionaly convergent (*2)
[tex] \sum b_n [/tex] is absolutly convergent (*3)
And I have seen this proof: [Proving [tex] \sum c_n[/tex] is conditionally convergent]
From (*1) and (*2) [tex]\Rightarrow[/tex] [tex] \sum c_n[/tex] its convergent [this one I understand, basic properties of series]
But now they do something like this: [proof by contradiction]
Supose [tex] \sum |c_n| [/tex] is convergent
so [tex] |a_n|\leq|c_n-b_n|\leq|c_n|+|b_n| [/tex] (How they "jump" to this conclusion?!)
and now the use comparison test to show that [tex]\sum c_n[/tex] is conditionally convergent. [No problems from here]
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