# Thread: Analize convergence of an infinite series

1. not that fast..

look what you have posted before my first post here..

2. Originally Posted by Moo
Why ?

$\sum (a_n+b_n)=a_1+b_1+a_2+b_2+\dots$

$\sum a_n+\sum b_n=a_1+a_2+\dots+b_1+b_2+\dots$

rearranging the terms won't change the convergence(...) or divergence, so :

$\sum a_n+\sum b_n=a_1+b_1+a_2+b_2+\dots=\sum (a_n+b_n)$

The most important thing here, imo, is that the domain of n for the two sums is the same, isn't it ?
(...) if $\sum (a_n+b_n)$ is absolutely convergent..

3. example again:

$a_n = \frac{1}{n^2}$
$b_n = \frac{1}{n}$

$\sum \frac{1}{n^2}$ is convergent.
$\sum \frac{1}{n}$ is divergent.

according to this:
Originally Posted by Moo
The sum of a divergent series and a convergent series is indeed divergent
$\sum \frac{1}{n^2} - \sum \frac{1}{n}$ must be divergent..

how about $\sum \frac{1-n}{n^2} = \sum \left(\frac{1}{n^2} - \frac{1}{n}\right)$? will that be divergent? check it..

4. what i mean there is that your statement in red is only true for absolutely convergent series.. and not generally..

5. Actually, I'm completely confused

I'll think about it this afternoon.

6. Cool, now I'm confused too.

7. no need to be confused.
what i am saying is that the equality
$\sum (a_b + b_n) = \sum a_n + \sum b_n$
is always true if both $\sum a_n$ and $\sum b_n$ are convergent.
and it is not true if at least 1 is divergent..

8. Oh yeah, that's right. Thanks

9. Originally Posted by Moo
Actually, I'm completely confused

I'll think about it this afternoon.
For you to be able to manipulate a series in the manner you are referring to it must be Cesaro Summable, and one of those conditions is that the series is convergent.

An example where rearanging is not permissable is the following

$x=1-1+1-1\cdots$

$\Rightarrow{x=1-(1-1+\cdots)}$

$\Rightarrow{x=1-x}$

$\Rightarrow{x=\frac{1}{2}}$

So

$\sum_{n=0}^{\infty}(-1)^n=\frac{1}{2}$

10. Originally Posted by Moo
$\sum (a_n+b_n)=a_1+b_1+a_2+b_2+\dots$

$\sum a_n+\sum b_n=a_1+a_2+\dots+b_1+b_2+\dots$

rearranging the terms won't change the convergence or divergence, so :

$\sum a_n+\sum b_n=a_1+b_1+a_2+b_2+\dots=\sum (a_n+b_n)$

The most important thing here, imo, is that the domain of n for the two sums is the same, isn't it ?
If $a_n$ and $b_n$ are convergent then $\Sigma_{n=1}^{\infty}(a_n+b_n) = \Sigma_{n=1}^{\infty} a_n + \Sigma_{n=1}^{\infty}b_n$ (no absolute convergence is used here).

If $a_n$ or $b_n$ are divergent then it does not even make sense to write $\Sigma_{n=1}^{\infty}a_n + \Sigma_{n=1}^{\infty}b_n$. Because what is that even supposed to mean?

For you to be able to manipulate a series in the manner you are referring to it must be Cesaro Summable, and one of those conditions is that the series is convergent.

An example where rearanging is not permissable is the following
There is no rearrangment going on over here (in what Moo did). Just term by term addition.

Let $a_n$ and $b_n$ be convergent series. Then $s_n = a_1+...+a_n$ and $t_n = b_1+...+b_n$ are convergent sequences. Now if $s_n$ and $t_n$ are convergent it means $s_n+t_n$ is a convergent. Furthermore, $\lim ~ (s_n+t_n) = \lim ~ s_n + \lim ~ t_n$. By definition, $\lim ~ s_n = \Sigma_{n=1}^{\infty}a_n$ and $\lim ~ t_n = \Sigma_{n=1}^{\infty}b_n$. And $\lim ~ (s_n+t_n) = \Sigma_{n=1}^{\infty} (a_n+b_n)$. Thus, $\Sigma_{n=1}^{\infty}(a_n+b_n) = \Sigma_{n=1}^{\infty} a_n + \Sigma_{n=1}^{\infty}b_n$.

11. Ok, I've talked to someone. Things are clearer

- $\sum (a_n+b_n)=\sum a_n+\sum b_n$ if both are convergent infinite series, or if both are finite series.
- the sentence for "sum of a divergent+a convergent is divergent" is true according to this person. Can anyone confirm or infirm ?

12. Originally Posted by Moo
- the sentence for "sum of a divergent+a convergent is divergent" is true according to this person. Can anyone confirm or infirm ?
That is easy. If you combine something that is finite and infinite you get something that is infinite. So what is the problem?

Just joking! That is how they "prove" in physics courses. Say $\Sigma a_n$ is convergent and $\Sigma b_n$ is divergent. And assume that $\Sigma (a_n+b_n)$ is convergent. Then it would mean $\Sigma (a_n+b_n) - a_n = \Sigma b_n$ is convergent. A contradiction!

13. Originally Posted by ThePerfectHacker
Just joking! That is how they "prove" in physics courses. Say $\Sigma a_n$ is convergent and $\Sigma b_n$ is divergent. And assume that $\Sigma (a_n+b_n)$ is convergent. Then it would mean $\Sigma (a_n+b_n) - a_n = \Sigma b_n$ is convergent. A contradiction!
EDIT:i misread your proof last night.. so probably, i should erase my previous comment..

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