the formula $$\sum_{i=1}^{n}\frac{1}{2^i}= \frac{1}{2}( 1+ \frac{1}{2} + \frac{1}{2^2} + \ldots + \frac{1}{2^{n-1}}) = \frac{1}{2} \cdot \frac{1-\frac{1}{2}^{n}}{1-\frac{1}{2}}  = 1-\frac{1}{2^n}$$ can be used directly for infinite $n$, too. For example, if $n=\G1$ then $$\sum_{i=1}^{\mbox{\small{\G1}}}\frac{1}{2^i}=1-\frac{1}{2^{\mbox{\tiny{\G1}}}},$$ where $\frac{1}{2^{\mbox{\tiny{\G1}}}}$ is infinitesimal. Thus, the traditional answer $\sum_{i=1}^{\infty}\frac{1}{2^i}=1$ was just a finite approximation to our more precise result using infinitesimals.

In order to show the potential of the new approach for the work with divergent series with alternate signs, we start from the famous series $$S_5 = 1-1+1-1+1-1+\ldots$$ In literature, there exist many approaches giving different answers regarding the value of this series (see [@Knopp]). All of them use various notions of average to calculate the series. However, the notions of the sum and an average are different. In our approach, we do not appeal to an average and calculate the required sum directly. To do this we should indicate explicitly the number of items, $k$, in the sum. Then S\_5(k)=\_[k ]{} = {

[ll]{} 0, & k=2n,\
1, & k=2n+1,\

. \[Riemann27\] where $k$ can be finite or infinite. For example, $S_5(\G1)=0$ because 1 is even (see [@informatica]). Analogously, $S_5(\G1-1)=S_5(2\G1+1)=1$ because both $\G1-1$ and $2\G1+1$ are odd.

It is important to emphasize that the answers obtained 