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
??
Best Answer
Consider the following
$$\sum_{n=1}^\infty \frac 1n =\infty,\qquad \sum_{n=1}^\infty \frac 1{2n}=\infty,\qquad \sum_{n=1}^\infty \frac 1{2n-1}=\infty,\qquad \sum_{n=1}^\infty \frac{(-1)^{n-1}}n=\ln(2).$$
In short: the sum of the reciprocals of all numbers diverges (harmonic series). The sum of the reciprocals of even/odd numbers diverges. But if you sum up even and odd reciprocals with alternating signs, it converges.
So here you have a specific infinite sum that seems to converge: $$1-\frac12+\frac13-\frac14+\cdots =\ln(2).$$
Of course, if you just rearrange finitely many of the summands, you will end up with the same sum. But if you shuffle up all the numbers, you can get everywhere with your limit. I think this is done in any proof of Riemann's rearrangement theorem, but let me line out the proof on this example. Lets say you want the rearranged sum to converge to $\pi$ (for fun). Then take some of the positive terms (the odd reciprocals) and add enough of them up until you are just greater than $\pi$: $$1+\frac13+\frac15+\cdots>\pi.$$ You can do this, as we know that the sum of the odd reciprocals diverges. In the next step, only take the negative (even) reciprocals and subtract them from your sum until you are just below $\pi$. Again you can do this, because the sum of the even reciprocals diverges. Now again take positiv terms, then negatives, then positives, and so on. In "the end" you will have used all the terms of the original sum, but you rearranged them in a way, so that they converge to $\pi$. And there is nothing special about $\pi$, so you can use this method to converge to anything, including $\pm\infty$.
Here a description of how to rearrange the sum to make it divergent, e.g. divergent to $\infty$. Sum up enough positive (odd) terms to make the sum greater than $1$. Add only a single negative term. Add positive terms until the sum exceed $2$. Add a single negative term. Add positive terms to exceed $3$, ... and so on. You will exceed any natural number, hence diverge to $\infty$.