Suppose that $a_0=1$ and that $a_{n+1}=a_n+e^{-a_n}$ for $n=0,1,2,\dots.$ Does $a_n-\log n$ have a finite limit as $n\to\infty?$ (Here $\log n=\log_en=\ln n.$)
1. **Define the sequence and the goal:**
   We are given the sequence \( \{a_n\} \) defined by \( a_0 = 1 \) and \( a_{n+1} = a_n + e^{-a_n} \) for \( n = 0, 1, 2, \ldots \). We need to determine if \( a_n - \log n \) has a finite limit as \( n \to \infty \).

2. **Introduce the sequence \( \{b_n\} \):**
   Let \( b_n = a_n - \ln n \). We aim to show that \( \{b_n\} \) converges. Note that \( b_1 = a_1 - \ln 1 = a_1 > 0 \).

3. **Substitute \( a_n = b_n + \ln n \) into the recurrence relation:**
   \[
   a_{n+1} = a_n + e^{-a_n} \implies b_{n+1} + \ln(n+1) = b_n + \ln n + e^{-(b_n + \ln n)}
   \]
   Simplifying, we get:
   \[
   b_{n+1} = b_n + \ln \left( \frac{n}{n+1} \right) + \frac{e^{-b_n}}{n}
   \]

4. **Show that \( b_n \geq 0 \) for all \( n \):**
   - Since \( e^x \geq x + 1 \) for all real \( x \), we have \( e^{-b_n} \geq 1 - b_n \).
   - For \( x = \frac{1}{n} \), \( e^{\frac{1}{n}} \geq 1 + \frac{1}{n} \), which rearranges to \( \frac{1}{n} + \ln \left( \frac{n}{n+1} \right) \geq 0 \).

   Therefore,
   \[
   b_{n+1} = b_n + \ln \left( \frac{n}{n+1} \right) + \frac{e^{-b_n}}{n} \geq b_n + \ln \left( \frac{n}{n+1} \right) + \frac{1 - b_n}{n}
   \]
   Simplifying further,
   \[
   b_{n+1} \geq b_n + \ln \left( \frac{n}{n+1} \right) + \frac{1}{n} - \frac{b_n}{n} \geq b_n \left( 1 - \frac{1}{n} \right)
   \]
   Since \( b_1, b_2 > 0 \), it follows by induction that \( b_n > 0 \) for all \( n \).

5. **Bound \( b_{n+1} \) from above:**
   \[
   b_{n+1} = b_n + \ln \left( \frac{n}{n+1} \right) + \frac{e^{-b_n}}{n} \leq b_n + \frac{1}{n} - \ln \left( 1 + \frac{1}{n+1} \right)
   \]
   Using the approximation \( \ln(1 + x) \approx x - \frac{x^2}{2} \) for small \( x \),
   \[
   b_{n+1} \leq b_n + \frac{1}{n} - \left( \frac{1}{n+1} - \frac{1}{2(n+1)^2} \right) = b_n + \frac{1}{2n^2} - O \left( \frac{1}{n^3} \right)
   \]
   Thus, there exists a positive constant \( c \) such that \( b_{n+1} \leq b_n + \frac{c}{n^2} \) for all \( n \).

6. **Sum the differences to show convergence:**
   For \( n > 1 \) and \( k > 0 \),
   \[
   b_{n+k} - b_n = \sum_{i=n}^{n+k-1} (b_{i+1} - b_i) \leq \sum_{i=n}^{n+k-1} \frac{c}{i^2} < \sum_{i=n}^{\infty} \frac{c}{i^2} < c \int_{i=n-1}^{\infty} \frac{1}{i^2} \, di = \frac{c}{n-1}
   \]

7. **Conclude that \( \{b_n\} \) is bounded and converges:**
   Since \( \{b_n\} \) is bounded from below, it has a limit inferior \( L \). Let \( \epsilon \) be any positive real number. There exists some \( N \) such that for all \( n > N \), \( b_n > L - \epsilon \). There also exist arbitrarily large \( M \) with \( b_M < L + \frac{\epsilon}{2} \). Take any such \( M \) for which \( \frac{c}{M-1} < \frac{\epsilon}{2} \). Then for all \( n > \max(M, N) \),
   \[
   b_n > L - \epsilon \quad \text{and} \quad b_n < b_M + (b_n - b_M) < (L + \frac{\epsilon}{2}) + \frac{\epsilon}{2} = L + \epsilon
   \]
   Thus, \( |b_n - L| < \epsilon \), implying \( \{b_n\} \) converges to \( L \).

8. **Show that \( L = 0 \):**
   Since \( b_n \geq 0 \), \( L \geq 0 \). Suppose for contradiction that \( L > 0 \). If we take \( \epsilon = \frac{L}{2} \), then there is some \( N \) such that whenever \( n > N \), \( b_n > \epsilon \). Now,
   \[
   b_{n+1} = b_n + \ln \left( \frac{n}{n+1} \right) + \frac{e^{-b_n}}{n} = b_n - \ln \left( 1 + \frac{1}{n} \right) + \frac{e^{-b_n}}{n}
   \]
   Using the approximation \( \ln(1 + x) \approx x - \frac{x^2}{2} \),
   \[
   b_{n+1} = b_n - \frac{1}{n} + O \left( \frac{1}{n^2} \right) + \frac{e^{-b_n}}{n} = b_n + \frac{e^{-b_n} - 1}{n} + O \left( \frac{1}{n^2} \right)
   \]
   Since \( e^{-\epsilon} - 1 < 0 \),
   \[
   b_{n+1} \leq b_n + O \left( \frac{1}{n^2} \right) + \frac{e^{-\epsilon} - 1}{n}
   \]
   Summing these from \( N \) to \( N + m \) for some \( m \),
   \[
   b_{N+m} \leq b_N + O \left( \frac{1}{N+m} \right) + (e^{-\epsilon} - 1) \sum_{i=N}^{N+m} \frac{1}{i}
   \]
   This is impossible since \( e^{-\epsilon} - 1 < 0 \) and \( \sum_{i=N}^{N+m} \frac{1}{i} \) diverges as \( m \) goes to infinity. Therefore, \( L = 0 \).

\(\blacksquare\)

The final answer is \( \boxed{ \lim_{n \to \infty} (a_n - \log n) = 0 } \)