in all the examples considered above (including the latter related to the divergent series $S_5(k)$ with alternate signs) do not depend on the way the items in the entire sum are re-arranged. In fact, since we always know the exact infinite number of items in the sum and the order of alternating the signs is clearly defined, we know also the exact number of positive and negative items in the sum.

Suppose, for instance, that we want to re-arrange the items in the sum $S_5(\G1)$ in the following way $$1+1-1+1+1-1+1+1-1+\ldots$$ However, we know that the sum $S_5(\G1)$ has $\G1$ items and that grossone is even. This means that in the sum there are $\frac{\G1}{2}$ positive and $\frac{\G1}{2}$ negative items. As a result, the re-arrangement considered above can continue only until the positive items will not finish and then it will be necessary to continue to add only negative numbers. More precisely, we have S\_5(1)=\_\_=0, \[Riemann28\] where the result of the first part in this re-arrangement is calculated as $(1+1-1)\cdot\frac{\mbox{\tiny{\G1}}}{4}=\frac{\mbox{\tiny{\G1}}}{4}$ and the result of the second part is equal to $-\frac{\mbox{\tiny{\G1}}}{4}$.

Note that the record (\[Riemann28\]) is a re-arrangement of the sum introduced in (\[Riemann27\]) where the order of the alternating 