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!