Let $ f:[0,1]\longrightarrow [0,1] $ be a nondecreasing function. Prove that the sequence
$$ \left( \int_0^1 \frac{1+f^n(x)}{1+f^{1+n} (x)} \right)_{n\ge 1} $$
is convergent and calculate its limit.
1. **Define the sequence and the function:**
   Let \( f: [0,1] \to [0,1] \) be a nondecreasing function. We need to prove that the sequence 
   \[
   \left( \int_0^1 \frac{1+f^n(x)}{1+f^{n+1}(x)} \, dx \right)_{n \ge 1}
   \]
   is convergent and calculate its limit.

2. **Pointwise behavior of the integrand:**
   Consider the integrand \( \frac{1+f^n(x)}{1+f^{n+1}(x)} \). Since \( f \) is nondecreasing and maps \([0,1]\) to \([0,1]\), we have \( 0 \le f(x) \le 1 \) for all \( x \in [0,1] \).

3. **Pointwise limit:**
   For any \( x \in [0,1] \), as \( n \to \infty \), \( f^n(x) \) converges to a limit \( L \in [0,1] \). If \( f(x) < 1 \), then \( f^n(x) \to 0 \). If \( f(x) = 1 \), then \( f^n(x) = 1 \) for all \( n \). Therefore, pointwise, we have:
   \[
   \frac{1+f^n(x)}{1+f^{n+1}(x)} \to 1.
   \]

4. **Uniform bound:**
   We need to show that the integrand is uniformly bounded. Note that:
   \[
   \left| \frac{1+f^n(x)}{1+f^{n+1}(x)} \right| \le 2.
   \]
   This is because \( 0 \le f(x) \le 1 \), so \( 1 \le 1+f^n(x) \le 2 \) and \( 1 \le 1+f^{n+1}(x) \le 2 \).

5. **Dominated Convergence Theorem:**
   By the Dominated Convergence Theorem, since \( \frac{1+f^n(x)}{1+f^{n+1}(x)} \) converges pointwise to 1 and is uniformly bounded by 2, we can interchange the limit and the integral:
   \[
   \lim_{n \to \infty} \int_0^1 \frac{1+f^n(x)}{1+f^{n+1}(x)} \, dx = \int_0^1 \lim_{n \to \infty} \frac{1+f^n(x)}{1+f^{n+1}(x)} \, dx = \int_0^1 1 \, dx = 1.
   \]

6. **Conclusion:**
   Therefore, the sequence 
   \[
   \left( \int_0^1 \frac{1+f^n(x)}{1+f^{n+1}(x)} \, dx \right)_{n \ge 1}
   \]
   is convergent and its limit is 1.

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