Problem 1553

1. Define the sequence and initial conditions:
The sequence is defined by $a_1 = 2022$ and $a_{n+1} = a_n + e^{-a_n}$ for $n \geq 1$.

2. **Behavior of the sequence $a_n$:**
To understand the behavior of $a_n$, we note that $e^{-a_n}$ is a very small positive number since $a_n$ is large. Therefore, $a_{n+1}$ is slightly larger than $a_n$, implying that $a_n$ is strictly increasing.

3. **Convergence of $a_n - \log(n)$:**
We need to show that $a_n - \log(n)$ converges. Consider the sequence $b_n = a_n - \log(n)$. We have:
$$a_{n+1} = a_n + e^{-a_n}$$
Subtracting $\log(n+1)$ from both sides, we get:
$$a_{n+1} - \log(n+1) = a_n - \log(n+1) + e^{-a_n}$$
Using the approximation $\log(n+1) \approx \log(n) + \frac{1}{n}$ for large $n$, we have:
$$a_{n+1} - \log(n+1) \approx a_n - \log(n) - \frac{1}{n} + e^{-a_n}$$
Therefore,
$$b_{n+1} \approx b_n - \frac{1}{n} + e^{-a_n}$$
Since $e^{-a_n}$ is very small, $b_n$ converges to some limit $L$.

4. **Choosing $r$:**
We take $r = \frac{1}{\log(10)}$. Then, we consider the sequence $\{ ra_{10^n} \}$.

5. **Behavior of $ra_{10^n}$:**
We have:
$$ra_{10^n} = r(a_{10^n} - \log(10^n) + \log(10^n)) = r(a_{10^n} - n \log(10)) + rn \log(10)$$
Since $a_{10^n} - n \log(10)$ converges, let it converge to $L$. Then,
$$ra_{10^n} = rL + n$$

6. Convergence of the fractional part:
The fractional part $\{ ra_{10^n} \}$ is given by:
$$\{ ra_{10^n} \} = \{ rL + n \}$$
Since $rL$ is a constant and $n$ is an integer, the sequence $\{ ra_{10^n} \}$ converges to $\{ rL \}$.

Therefore, we have shown that there exists a positive real number $r$ such that the sequence $\{ ra_1 \}, \{ ra_{10} \}, \{ ra_{100} \}, \ldots$ converges.

$\blacksquare$