Define a sequence $(a_n)$ by $a_0 =0$ and $a_n = 1 +\sin(a_{n-1}-1)$ for $n\geq 1$. Evaluate
$$\lim_{n\to \infty} \frac{1}{n} \sum_{k=1}^{n} a_k.$$
1. Define the sequence \( (a_n) \) by \( a_0 = 0 \) and \( a_n = 1 + \sin(a_{n-1} - 1) \) for \( n \geq 1 \).
2. To simplify the sequence, let \( b_n = a_n - 1 \). Then, we have:
   \[
   b_0 = a_0 - 1 = -1
   \]
   and for \( n \geq 1 \):
   \[
   b_n = a_n - 1 = 1 + \sin(a_{n-1} - 1) - 1 = \sin(a_{n-1} - 1) = \sin(b_{n-1})
   \]
3. We need to evaluate:
   \[
   \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} a_k
   \]
   Using the substitution \( b_n = a_n - 1 \), we rewrite the sum:
   \[
   \frac{1}{n} \sum_{k=1}^{n} a_k = \frac{1}{n} \sum_{k=1}^{n} (b_k + 1) = \frac{1}{n} \sum_{k=1}^{n} b_k + \frac{1}{n} \sum_{k=1}^{n} 1 = \frac{1}{n} \sum_{k=1}^{n} b_k + 1
   \]
4. We need to find:
   \[
   \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} b_k
   \]
5. Note that \( b_n = \sin(b_{n-1}) \). Since \( |\sin(x)| < |x| \) for all \( x \neq 0 \), the sequence \( (b_n) \) is decreasing in absolute value and converges to 0. Therefore, \( \lim_{n \to \infty} b_n = 0 \).
6. By the Stolz-Cesaro theorem, if \( \lim_{n \to \infty} b_n = 0 \), then:
   \[
   \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} b_k = 0
   \]
7. Therefore:
   \[
   \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} a_k = \lim_{n \to \infty} \left( \frac{1}{n} \sum_{k=1}^{n} b_k + 1 \right) = 0 + 1 = 1
   \]

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