Conditional Convergence and the Comparison Test: A Proof by Contradiction

[estro]

Series convergence "by Parts

Supose:

\sum c_n = \sum (a_n+b_n) (*1)

\sum a_n is conditionaly convergent (*2)

\sum b_n is absolutly convergent (*3)

And I have seen this proof: [Proving \sum c_n is conditionally convergent]

From (*1) and (*2) \Rightarrow \sum c_n its convergent [this one I understand, basic properties of series]

But now they do something like this: [proof by contradiction]

Supose \sum |c_n| is convergent

so |a_n|\leq|c_n-b_n|\leq|c_n|+|b_n| (How they "jump" to this conclusion?!)

and now the use comparison test to show that \sum c_n is conditionally convergent. [No problems from here]

 

[LCKurtz]

It's easy:

c_n = a_n + b_n
a_n = c_n - b_n
|a_n| = |c_n - b_n| ≤ |c_n| + |b_n|

This would imply the a_n series converges absolutely, which is a contradiction.

 

[estro]

I don't understand this:

If\ \sum c_n = \sum (a_n+b_n)\ convergent,

why\ c_n=a_n+b_n.

Don't understand the theory.

 

[LCKurtz]

You are given two series, ∑a_n and ∑b_n and examining their sum

∑a_n + ∑b_n = ∑(a_n+b_n)

to see if it is conditionally convergent. It is just a convenience to call the term

a_n+b_n on the right side c_n. There is nothing to prove about that.

 

[estro]

Oh, bad idea to do such things in a book. (confusing)

Thanks Again!