A strip of width $1$ is to be divided by rectangular panels of common width $1$ and denominations long $a_1$, $a_2$, $a_3$, $. . .$ be paved without gaps ($a_1 \ne 1$). From the second panel on, each panel is similar but not congruent to the already paved part of the strip. When the first $n$ slabs are laid, the length of the paved part of the strip is $sn$. Given $a_1$, is there a number that is not surpassed by any $s_n$? The accuracy answer has to be proven.
1. **Understanding the Problem:**
   - We have a strip of width 1.
   - The strip is divided into rectangular panels of width 1 and lengths \(a_1, a_2, a_3, \ldots\).
   - The panels are laid without gaps.
   - From the second panel onwards, each panel is similar but not congruent to the already paved part of the strip.
   - After laying the first \(n\) slabs, the length of the paved part of the strip is \(s_n\).
   - We need to determine if there is a number that is not surpassed by any \(s_n\).

2. **Analyzing the Panels:**
   - The first panel has length \(a_1\).
   - The second panel is similar but not congruent to the first panel, meaning it is a scaled version of the first panel.
   - Let the scaling factor be \(k\). Then the length of the second panel is \(k \cdot a_1\).
   - The third panel is similar to the first two panels, so its length is \(k^2 \cdot a_1\), and so on.

3. **Generalizing the Lengths:**
   - The length of the \(n\)-th panel is \(k^{n-1} \cdot a_1\).
   - The total length of the paved part of the strip after \(n\) panels is:
     \[
     s_n = a_1 + k \cdot a_1 + k^2 \cdot a_1 + \cdots + k^{n-1} \cdot a_1
     \]

4. **Summing the Series:**
   - The series \(a_1 + k \cdot a_1 + k^2 \cdot a_1 + \cdots + k^{n-1} \cdot a_1\) is a geometric series with the first term \(a_1\) and common ratio \(k\).
   - The sum of the first \(n\) terms of a geometric series is given by:
     \[
     s_n = a_1 \frac{1 - k^n}{1 - k} \quad \text{for} \quad k \neq 1
     \]

5. **Behavior of the Series:**
   - If \(0 < k < 1\), as \(n\) approaches infinity, \(k^n\) approaches 0.
   - Therefore, the sum \(s_n\) approaches:
     \[
     s_n \to a_1 \frac{1}{1 - k}
     \]
   - This means that there is a limit to the length of the paved part of the strip, which is \(a_1 \frac{1}{1 - k}\).

6. **Conclusion:**
   - Given \(a_1\), there is indeed a number that is not surpassed by any \(s_n\), which is \(a_1 \frac{1}{1 - k}\).

The final answer is \( \boxed{ a_1 \frac{1}{1 - k} } \).