# Series convergence by Parts

1. Jun 10, 2010

### 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]

Last edited: Jun 10, 2010
2. Jun 10, 2010

### LCKurtz

Re: Series convergence "by Parts

It's easy:

cn = an + bn
an = cn - bn
|an| = |cn - bn| ≤ |cn| + |bn|

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

3. Jun 10, 2010

### estro

Re: Series convergence "by Parts

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.

4. Jun 10, 2010

### LCKurtz

Re: Series convergence "by Parts

You are given two series, ∑an and ∑bn and examining their sum

∑an + ∑bn = ∑(an+bn)

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

an+bn on the right side cn. There is nothing to prove about that.

5. Jun 10, 2010

### estro

Re: Series convergence "by Parts

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

Thanks Again!