of positive and negative items has been defined. Thanks to this order and to the knowledge of the whole number of items in the sum (\[Riemann27\]) we were able to calculate the number of negative and positive items. Obviously, if we consider another sum where the order of the alternating of the signs and the number of the items are given in a different way then the result can be also different. For instance, if we have the sum $$S_6(\G1)=\underbrace{1+1-1+1+1-1+\ldots }_{ \G1 \mbox{ items} }$$ then we have $\frac{2\mbox{\tiny{\G1}}}{3}$ positive items and $\frac{\mbox{\tiny{\G1}}}{3}$ negative items that gives the result $S_6(\G1)=\frac{\mbox{\tiny{\G1}}}{3}$.

Let us consider now the following divergent series $$S_7=  1 -2   + 3 -4 + \ldots$$ The corresponding sum $S_7(k)$ can be easily calculated as the difference of two arithmetic progressions after we have fixed the infinite number of items, $k$, in it. Suppose that $k=\G1$. Then it follows $$S_7(\G1) = \underbrace{1 -2   + 3 -4 + \ldots - (\G1-2) + (\G1-1)
- \G1}_{ \G1
 \mbox{ items}} =$$ $$\underbrace{(1     + 3   + 5 + \ldots + (\G1-3) + (\G1-1)
)}_{\frac{\mbox{\tiny{\G1}}}{2}\mbox{ items}} - \underbrace{( 2 +
4 + 6 + \ldots + (\G1-2) + \G1
)}_{\frac{\mbox{\tiny{\G1}}}{2}\mbox{ items}} =$$ - = = 