Define the sequence $a_0,a_1,\dots$ inductively by $a_0=1$, $a_1=\frac{1}{2}$, and 
\[a_{n+1}=\dfrac{n a_n^2}{1+(n+1)a_n}, \quad \forall n \ge 1.\]
Show that the series $\displaystyle \sum_{k=0}^\infty \dfrac{a_{k+1}}{a_k}$ converges and determine its value.

[i]Proposed by Christophe Debry, KU Leuven, Belgium.[/i]
1. We start by analyzing the given sequence \(a_n\). The sequence is defined as:
   \[
   a_0 = 1, \quad a_1 = \frac{1}{2}, \quad \text{and} \quad a_{n+1} = \frac{n a_n^2}{1 + (n+1) a_n} \quad \forall n \ge 1.
   \]

2. We need to show that the series \(\sum_{k=0}^\infty \frac{a_{k+1}}{a_k}\) converges and determine its value.

3. First, we consider the expression \(na_n - (n+1)a_{n+1}\):
   \[
   na_n - (n+1)a_{n+1} = na_n - \frac{n(n+1) a_n^2}{1 + (n+1)a_n}.
   \]

4. Simplifying the right-hand side:
   \[
   na_n - \frac{n(n+1) a_n^2}{1 + (n+1)a_n} = \frac{n a_n (1 + (n+1)a_n) - n(n+1) a_n^2}{1 + (n+1)a_n} = \frac{n a_n}{1 + (n+1)a_n}.
   \]

5. Since \(\frac{n a_n}{1 + (n+1)a_n} > 0\), the sequence \((na_n)_{n \ge 1}\) is decreasing and bounded below by 0. Therefore, it converges to some limit \(\ell \ge 0\).

6. Passing to the limit as \(n \to \infty\):
   \[
   0 = \lim_{n \to \infty} \left(na_n - (n+1)a_{n+1}\right) = \lim_{n \to \infty} \frac{n a_n}{1 + (n+1)a_n}.
   \]

7. Simplifying the limit:
   \[
   \lim_{n \to \infty} \frac{n a_n}{1 + (n+1)a_n} = \lim_{n \to \infty} \frac{n a_n}{1 + \frac{n+1}{n} (n a_n)} = \lim_{n \to \infty} \frac{n a_n}{1 + (n a_n + a_n)}.
   \]

8. Since \(na_n\) converges to \(\ell\), we have:
   \[
   \frac{\ell}{1 + \ell} = 0 \implies \ell = 0.
   \]

9. We also have:
   \[
   \frac{n a_n}{1 + (n+1)a_n} = \frac{a_{n+1}}{a_n}.
   \]

10. By telescoping the series:
    \[
    \sum_{k=0}^{n-1} \frac{a_{k+1}}{a_k} = \frac{a_1}{a_0} + a_1 - na_n.
    \]

11. Taking the limit as \(n \to \infty\):
    \[
    \sum_{k=0}^\infty \frac{a_{k+1}}{a_k} = \lim_{n \to \infty} \sum_{k=0}^{n-1} \frac{a_{k+1}}{a_k} = \frac{a_1}{a_0} + a_1 - \lim_{n \to \infty} na_n.
    \]

12. Substituting the known values:
    \[
    \sum_{k=0}^\infty \frac{a_{k+1}}{a_k} = \frac{1/2}{1} + \frac{1}{2} - 0 = 1.
    \]

The final answer is \(\boxed{1}\).