infinite and infinitesimal numbers allows us to take into account infinite $n$ and the value $x^{n+1}$ being infinitesimal for a finite $x$. Moreover, we can calculate $Q_n$ for infinite and finite values of $n$ and $x=1$, because in this case we have just $$Q_n = \underbrace{1+1+1+\ldots+1}_{n+1 \mbox{ items}} = n+1.$$

Let us illustrate the obtain results by some examples. In the first of them we consider the divergent series $$1 + 3 + 9 + \ldots = \sum_{i=0}^{\infty} 3^i.$$ In the new language, to define a sum it is necessary to fix the number of items, $n$, in it and $n$ can be finite or infinite. For example, for the infinite $n=\G1^2$ we obtain $$\sum_{i=0}^{\G1^2} 3^i = 1 + 3 + 9 + \ldots +
3^{\mbox{\tiny{\G1}}^2}= \frac{1-3^{\mbox{\tiny{\G1}}^2+1}}{1-3} =
0.5(3^{\mbox{\tiny{\G1}}^2+1}-1).$$ Analogously, for $n=\G1^2+1$ we obtain $$1 + 3 + 9 + \ldots + 3^{\mbox{\tiny{\G1}}^2} +
3^{\mbox{\tiny{\G1}}^{2}+1}= 0.5(3^{\mbox{\tiny{\G1}}^2+2}-1).$$ If we now find the difference between the two sums $$0.5(3^{\mbox{\tiny{\G1}}^2+2}-1) -0.5(3^{\mbox{\tiny{\G1}}^2+1}-1)
= 3^{\mbox{\tiny{\G1}}^2+1} (0.5\cdot3 - 0.5) =
3^{\mbox{\tiny{\G1}}^{2}+1}$$ we obtain the newly added item $3^{\mbox{\tiny{\G1}}^2+1}$.

Let us now consider the series $\sum_{i=1}^{\infty}\frac{1}{2^i}$. It is well known that it converges to one. However, we are able to give a more precise answer. In fact, 