$S_2(n) - S_1(k)$ (analogously, the expression $ \frac{S_1(k)}{S_2(n)} $ can be calculated). If, for instance, $k=5\G1$ and $n=\G1$ we obtain $S_1(5\G1)=35\G1$, $S_2(\G1)=3\G1$ and it follows $$S_2(\G1) - S_1(5\G1)= 3\G1 - 35\G1 = - 32\G1< 0.$$ If $k=3\G1$ and $n=7\G1+2$ we obtain $S_1(3\G1)=21\G1$, $S_2(7\G1+2)=21\G1+6$ and it follows $$S_2(7\G1+2) -  S_1(3\G1)= 21\G1+6 - 21\G1 = 6.$$ It becomes also possible to calculate sums having an infinite number of infinite or infinitesimal items. Let us consider, for instance, sums $$S_3(l)=\underbrace{2\G1+2\G1+\ldots+2\G1}_{l \mbox{ items}},
\hspace{1cm}
S_4(m)=\underbrace{4\G1^{-1}+4\G1^{-1}+\ldots+4\G1^{-1}}_{m \mbox{
items}}.$$ For $l=m=0.5\G1$ it follows $S_3(0.5\G1)= \G1^2$ and $S_4(0.5\G1)=2$ (remind that $\G1\cdot\G1^{-1}=\G1^{0}=1$ (see (\[3.15.1\])). It can be seen from this example that it is possible to obtain finite numbers as the result of summing up infinitesimals.

The infinite and infinitesimal numbers allow us to calculate also arithmetic and geometric series with an infinite number of items. For the arithmetic progression, $a_{n} = a_{1} + (n - 1)d$, for both finite and infinite $n$ we have $$\sum_{i=1}^{n} a_{i} = \frac{n}{2}(a_{1} + a_{n}).$$ Then, for instance, the sum of all natural numbers from 1 to 1 can be calculated as follows 1+2+3+ …+ (1-1) + 1 = \_[i=1]{}\^[1]{} i = (1 + 1)= 0.51\^[2]{}0.51. \[Riemann1\] If we calculate now the 