How come rearrangement of a convergent series may not converge to the same value / or does not converge at all By looking at Riemann's Rearrangement theorem I wonder, How come a convergent series particular re-arrengement may not converge to the same value of the original series.
Isn't the below true?
Let $\phi : N \to N$ be a bijection. Where $N$ is the natural number set
A re-arrengement of a sequence $\sum a_n$ would be $\sum a_{\phi(n)}$
Then Give me an example of a series that have a rearrangement which does not converge or does not converge to the same value.
I mean... 4+1+2+2+9+6 = 1+2+2+6+4+9 = 24 
??

I think you need to understand the nature of a conditionally convergent series. I restrict the discussion to series whose terms are real numbers.
If a series $\sum a_{n}$ is conditionally convergent and we separate the positive terms of this series into $\sum b_{n}$ and negative terms into another series $\sum c_{n}$ then we have to understand following two things:


*

*$\lim_{n \to \infty} a_{n} = \lim_{n \to\infty}b_{n} = \lim_{n \to \infty}c_{n} = 0$

*$\sum b_{n}$ diverges to $\infty$ and $\sum c_{n}$ diverges to $-\infty$.


Consider the expression $$S(M,N) =\sum_{n=1}^{M} b_{n} + \sum_{n=1}^{N} c_{n}$$ and we can see that it is of the indeterminate form $\infty - \infty$ as $M, N$ tend to $\infty $ independently. Since the series on right diverge it is possible to take desired number of terms from $\sum b_{n}$ and $\sum c_{n}$ to add up to any particular sum (including $\pm\infty$). To add some details, suppose I want to achieve the sum $2$. Then I choose terms from $\sum b_{n}$ so that the sum of these chosen terms is greater than $2$ (this is possible because $\sum b_{n}$ diverges and we can exceed any number by choosing sufficient number of terms). Next I add terms from the series $\sum c_{n}$ (note the terms here are negative so that adding them effectively reduces the sum) so that the sum is less than $2$. Repeating this procedure ad infinitum I get a series whose sum is $2$.
Formally given any extended real number $S$, we can prove that there exist sequences $m_{k}, n_{k} $ taking positive integer values and tending to $\infty$ as $k\to\infty$ such that $S(m_{k}, n_{k}) \to S$ as $k\to\infty$. The above argument when formalized with all the details constitutes a proof of Riemann's rearrangement theorem.
