+ P^1 $, $P_1$ designates a positive quantity which is dependent on $n$ an $e$.

-   $S = \ \quad \frac{1 - e^n}{1 - e} \quad + \quad e^n [1 + e + \dots + \text{in inf.}]$

> The quantity $[1 + e + \dots + \text{in inf.}]$ is not to be regarded as identical to $S$, since the set of terms is not the same, rather here it is indisputably $n$ terms less than in $S$. [^15]

The following theorem and its proof demonstrates Bolzano’s way of adding infinite series and the sufficient condition of their order. He employs Galileo’s example of the series of natural numbers and the series of their squares (second powers), so $P$ and $S$ from the example above. The question isn’t which series has more terms - the multitude of terms is the same - but which sum of the series is greater. $$P = 1 + 2 + 3 + 4 + \dots + \text{in inf.}$$ $$S = 1 + 4 + 9 + 16 + \dots + \text{in inf.}$$

\[S&gt;P\] Let $P, S$ be the sums of infinite series from the example \[rady\]. Then

-   $S$ is greater than $P$, $S > 