<s>[INST] 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 
??
 [/INST] Here's a concrete example where rearrangement of a convergent series results in a series that converges to a different value.
Consider
$$
1 - 1 + \frac{1}{2} - \frac{1}{2} + \frac{1}{3} - \frac{1}{3}
+ \frac{1}{4} - \frac{1}{4} + \frac{1}{5} - \frac{1}{5} + \ldots
$$
The partial sums of this series alternate 0  between values of the form $1/n$, so it converges to zero.
Now consider
$$
1 + \frac{1}{2} - 1 + \frac{1}{3} + \frac{1}{4} - \frac{1}{2}
+ \frac{1}{5} + \frac{1}{6} - \frac{1}{3}
+ \frac{1}{7} + \frac{1}{8} - \frac{1}{4} + \ldots
$$
This is clearly a rearrangement of the first, and it's not too hard to see that you don't change the convergence behavior if you sum the terms in groups of 3:
$$
\left(1 + \frac{1}{2} - 1\right) + \left(\frac{1}{3} + \frac{1}{4} - \frac{1}{2}\right)
+ \left(\frac{1}{5} + \frac{1}{6} - \frac{1}{3}\right)
+ \left(\frac{1}{7} + \frac{1}{8} - \frac{1}{4}\right) + \ldots \\
= \frac{1}{2} + \frac{1}{12} +\frac{1}{30} + \frac{1}{56} + \ldots \\
$$
Now by standard methods you can check that this converges to some value greater than zero.  However, expanding each term as a difference shows that it actually converges to a known value.
$$
\frac{1}{2} + \frac{1}{12} +\frac{1}{30} + \frac{1}{56} + \ldots \\
=\left(1 - \frac{1}{2} \right) + \left(\frac{1}{3} - \frac{1}{4} \right)
+ \left(\frac{1}{5} - \frac{1}{6} \right)
+ \left(\frac{1}{7} - \frac{1}{8} \right) + \ldots \\ = \ln(2)
$$
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