1. Inversion Transformation:
Let $M$ and $N$ be the antipodes of $A_0$ in circles $\mathcal{A}$ and $\mathcal{B}$, respectively. Consider an inversion through the pole $A_0$ with power $\overline{A_0M} \cdot \overline{A_0N} = 4R \cdot r$. This inversion transforms the circles $\mathcal{A}$ and $\mathcal{B}$ into two lines $a$ and $b$ that are perpendicular to $A_0N$ and pass through $N$ and $M$, respectively.

2. Transformation of the Pappus Chain:
The sequence of circles $(C_n)$, which form a Pappus chain, are transformed into a sequence of congruent circles $(U_n)$ that are tangent to each other and to the lines $a$ and $b$. The contact points of the circles $U_n$ lie on the midline $\ell$ of $a$ and $b$.

3. Concyclic Points:
The points $A_n$ are concyclic on a circle $\mathcal{C}$ with radius $\varrho$. This circle $\mathcal{C}$ is the inverse image of the midline $\ell$.

4. **Calculation of Radius $\varrho$:**
Let $L$ be the midpoint of $MN$. We have:
$$\overline{A_0M} \cdot \overline{A_0N} = 4R \cdot r = 2\varrho \cdot A_0L = 2\varrho \cdot (R + r)$$
Solving for $\varrho$, we get:
$$\varrho = \frac{2Rr}{R + r}$$

5. Perimeter of the Inscribed Polygon:
Since the points $A_0, A_1, A_2, \ldots, A_n$ are inscribed in the circle $\mathcal{C}$, the perimeter of the polygon formed by these points is less than the circumference of $\mathcal{C}$. Therefore:
$$\sum_{n=1}^{\infty} |A_{n}A_{n+1}| < 2\pi \cdot \varrho$$

6. Final Inequality:
Substituting the value of $\varrho$:
$$2\pi \cdot \varrho = 2\pi \cdot \frac{2Rr}{R + r} = \frac{4\pi Rr}{R + r}$$
Hence, we have:
$$\sum_{n=1}^{\infty} |A_{n+1}A_n| < \frac{4\pi Rr}{R + r}$$

The final answer is $\boxed{ \sum_{n=1}^{\infty} |A_{n+1}A_n| < \frac{4 \pi Rr}{R+r} }$