following sum of infinitesimals where each item is 1 times less than the corresponding item of (\[Riemann1\]) $$\G1^{-1}+2\G1^{-1}+ \ldots + (\G1-1)\cdot\G1^{-1} +
\G1\cdot\G1^{-1} = \sum_{i=1}^{\G1} i\G1^{-1} =
\frac{\G1}{2}(\G1^{-1} + 1)= 0.5\G1^{1}0.5.$$ then, obviously, the obtained number, $0.5\G1^{1}0.5$, is 1 times less than the sum in (\[Riemann1\]). This example shows, particularly, that infinite numbers can also be obtained as the result of summing up infinitesimals.

Let us consider now the geometric series $\sum_{i=0}^{\infty}
x^i$. Traditional analysis proves that it converges to $\frac{1}{1-x}$ for $x$ such that $-1 < x < 1$. We are able to give a more precise answer for *all* values of $x$. To do this, we should fix the number of items in the sum. If we suppose that it contains $n$ items, where $n$ is finite or infinite, then Q\_n = \_[i=0]{}\^[n]{} x\^i = 1 + x + x\^2 + …+ x\^n. \[3.7.2.f\] By multiplying the left hand and the right hand parts of this equality by $x$ and by subtracting the result from (\[3.7.2.f\]) we obtain $$Q_n - xQ_n = 1-x^{n+1}$$ and, as a consequence, for all $x\neq 1$ the formula Q\_n = (1-x\^[n+1]{})(1-x)\^[-1]{} \[3.7.2.f.1\] holds for finite and infinite $n$. Thus, the possibility to express 