6. With these things thus established, there is the following

Theorem.

Let $s, s_1, s_2, \dots s_n$ be positive integers, the last of which, $s_n$, is greater than unity, and let the aggregates be formed

$c^{s-1} = γ \quad c^{s-1} d = c_1 \quad c^s d = d_1$

$c_1^{s_1-1} = γ_1 \quad c_1^{s_1-1} d_1 = c_2 \quad c_1^{s_1} d_1 = d_2$

$c_2^{s_2-1} = γ_2 \quad c_2^{s_2-1} d_2 = c_3 \quad c_2^{s_2} d_2 = d_3$

$\cdot \quad \cdot \quad \cdot \quad \cdot \quad \cdot \quad \cdot$

then the general form of $Π$ itself will be this:

$Π = c_1 c_2 \dots c_{n-1} γ_n γ_{n-1} \dots γ_2 γ_1 γ ,$

and from this it follows that

$\frac{a}{b} = \frac{1}{s + \frac{1}{s_1 + \frac{1}{s_2 \dots + \frac{1}{s_n}}}}$

7. In the second example, $Π = c^2 d c^2 d c^3 d c^2 d c^3 d c^2 d c^2$. From this it follows

that it must be $γ = c^2$, $c_1 = c^2 d$, therefore $s = 3$, $d_1 = c^3 d$, $γ_1 = c_1^{s_1-1}$ and

$Π = c_1 \cdot c_1 d_1 c_1 d_1 c_1 \cdot γ .$

Furthermore, it follows that $γ_1 = c_1$, therefore $s_1 = 2$, $c_2 = c_1 d_1$, $d_2 = c_1^2 d_1$, $γ_2 = c_2^{s_2-1}$ and
$Π = c_1 c_2 γ_1 γ$; finally, $γ_2 = c_2$, $s_2 = 2$, whence $Π = c_1 γ_2 γ_1 γ$ and

$\frac{a}{b} = \frac{1}{3 + \frac{1}{2 + \frac{1}{3}}} = \frac{7}{24} ,$

as it ought to be.

8. Thus it is evident:

I. how the principal part of the period is obtained from the expansion of the fraction $\frac{a}{b}$ into a continued fraction by means of the positive integers $s, s_1, s_2, \dots s_n$, the last of which, however, must be $> 1;$