Euler sum of a divergent series So I have a series $1+0+(-1)+0+(-1)+0+1+0+1+0+(-1)+...$
Is it correct to rearrange this as $1+0+(-1)+0+1+0+(-1)+0+1+0+(-1)+0...$
The second problem can be done as an Euler sum and the answer is $\frac{1}{2}$. In general, I know that absolutely convergent series can be rearranged but I'm not sure what the rule is for this case
 A: You know if it is absolutely convergent, you will have probably checked for conditional convergence somewhere. Therefore, you should be able to do it based of conditional convergence.
Turns out, if it converges conditionally , you can rearrange the terms in a permutation so the series converges, or diverges. Something good to look at would be Riemann Series Theorem.
Your series can be rearranged in one of the two following ways:
1.) $1+(1-1)+(1-1)+(1-1)+...$
2.) $(1-1)+(1-1)+(1-1)+...$
A: A series converges, by definition, if the sequence of partial sums $S_n$ converges to some number.   So, call your individual terms $a_n$,  the sequence of partial sums is $\displaystyle \sum _{k=1}^n a_k$.   If you look at your sequence of partial sums, you can see it doesn't converge to any number,  because it varies from 1 to 0 back to 1 infinitely often, therefore there is no $N$ such that $\forall n>N$,   you have a sum $S$ such that $|S_n -S|<\frac 1 2$ (picking one half as our epsilon challenge, for instance)
Another way you can tell that this sequence diverges is that in a necessary (But not sufficient) criterea for a series to converge is that the limit of the individual terms in the sequence goes to 0, and here once again you do not have that.
Now, some people will play around with alternate theories of mathematics that are NOT standard analysis that will assign a "sum" to normally divergent sequences.   But that's not what the default meaning for the value of an infinite sum.
A: One can only rearrange absolutely convergent series (positive terms convergent as well). For semi convergent series one can prove that rearranging can lead to any limit including making the series divergent. Numerically there is a very rich field around divergent series.
