Let $k\geqslant 2$ be an integer. Consider the sequence $(x_n)_{n\geqslant 1}$ defined by $x_1=a>0$ and $x_{n+1}=x_n+\lfloor k/x_n\rfloor$ for $n\geqslant 1.$ Prove that the sequence is convergent and determine its limit.
1. **Define the sequence and initial conditions:**
   The sequence \((x_n)_{n \geq 1}\) is defined by \(x_1 = a > 0\) and \(x_{n+1} = x_n + \left\lfloor \frac{k}{x_n} \right\rfloor\) for \(n \geq 1\).

2. **Analyze the difference between consecutive terms:**
   We observe that \(x_{n+1} - x_n = \left\lfloor \frac{k}{x_n} \right\rfloor\). Since \(\left\lfloor \frac{k}{x_n} \right\rfloor\) is an integer, \(x_{n+1} - x_n\) is also an integer.

3. **Behavior of the sequence for large \(n\):**
   - If \(x_n \leq k\), then \(\left\lfloor \frac{k}{x_n} \right\rfloor \geq 1\), so \(x_{n+1} > x_n\).
   - If \(x_n > k\), then \(\left\lfloor \frac{k}{x_n} \right\rfloor = 0\), so \(x_{n+1} = x_n\).

4. **Determine the limit:**
   - If \(x_n \leq k\), the sequence is strictly increasing.
   - Once \(x_n > k\), the sequence becomes constant because \(\left\lfloor \frac{k}{x_n} \right\rfloor = 0\).

5. **Convergence of the sequence:**
   Since the sequence is strictly increasing when \(x_n \leq k\) and becomes constant once \(x_n > k\), the sequence will eventually reach a value greater than \(k\) and stay constant thereafter.

6. **Finding the smallest element \(p\) such that \(p > k\):**
   - If \(a > k\), then the sequence is already constant with \(x_n = a\) for all \(n\).
   - If \(a \leq k\), we need to find the smallest integer \(p\) such that \(p > k\). This can be done by iterating the sequence until \(x_n > k\).

7. **Explicit calculation for \(p\):**
   - If \(a \leq k\), we find the smallest integer \(i\) such that \(a + i \leq k < a + i + 1\). This \(i\) is given by \(i = \left\lfloor k - a \right\rfloor\).
   - Therefore, \(p = a + i + 1 = a + 1 + \left\lfloor k - a \right\rfloor\).

The final answer is \( \boxed{ a + 1 + \left\lfloor k - a \right\rfloor } \).