Let $f: [0, 1] \to \mathbb{R}$ be a continuous strictly increasing function such that
\[ \lim_{x \to 0^+} \frac{f(x)}{x}=1. \]
(a) Prove that the sequence $(x_n)_{n \ge 1}$ defined by
\[ x_n=f \left(\frac{1}{1} \right)+f \left(\frac{1}{2} \right)+\cdots+f \left(\frac{1}{n} \right)-\int_1^n f \left(\frac{1}{x} \right) \mathrm dx \]
is convergent.
(b) Find the limit of the sequence $(y_n)_{n \ge 1}$ defined by
\[ y_n=f \left(\frac{1}{n+1} \right)+f \left(\frac{1}{n+2} \right)+\cdots+f \left(\frac{1}{2021n} \right). \]
### Part (a)

1. **Define the sequence and integral:**
   \[
   x_n = \sum_{k=1}^n f\left(\frac{1}{k}\right) - \int_1^n f\left(\frac{1}{x}\right) \, dx
   \]

2. **Analyze the behavior of \( f \left( \frac{1}{x} \right) \):**
   Since \( f \) is continuous and strictly increasing, and given that \( \lim_{x \to 0^+} \frac{f(x)}{x} = 1 \), we have:
   \[
   f\left(\frac{1}{x}\right) \sim \frac{1}{x} \quad \text{as} \quad x \to \infty
   \]

3. **Use the Riemann sum approximation:**
   The integral \( \int_1^n f\left(\frac{1}{x}\right) \, dx \) can be approximated by the Riemann sum:
   \[
   \int_1^n f\left(\frac{1}{x}\right) \, dx \approx \sum_{k=1}^n f\left(\frac{1}{k}\right) - f\left(\frac{1}{n}\right)
   \]
   This implies:
   \[
   f\left(\frac{1}{n}\right) < x_n < f(1)
   \]

4. **Bound the sequence \( x_n \):**
   Since \( f \left( \frac{1}{n} \right) \to 0 \) as \( n \to \infty \) and \( f(1) \) is a finite constant, \( x_n \) is bounded.

5. **Show that \( x_n \) is monotonic:**
   \[
   x_{n+1} = \sum_{k=1}^{n+1} f\left(\frac{1}{k}\right) - \int_1^{n+1} f\left(\frac{1}{x}\right) \, dx
   \]
   \[
   x_{n+1} = x_n + f\left(\frac{1}{n+1}\right) - \int_n^{n+1} f\left(\frac{1}{x}\right) \, dx
   \]
   Since \( f \left( \frac{1}{n+1} \right) < \int_n^{n+1} f\left(\frac{1}{x}\right) \, dx < f \left( \frac{1}{n} \right) \), we have:
   \[
   f\left(\frac{1}{n+1}\right) - \int_n^{n+1} f\left(\frac{1}{x}\right) \, dx < 0
   \]
   Thus, \( x_{n+1} < x_n \), meaning \( x_n \) is strictly decreasing.

6. **Conclude convergence:**
   Since \( x_n \) is bounded and strictly decreasing, by the Monotone Convergence Theorem, \( x_n \) converges.

### Part (b)

1. **Define the sequence \( y_n \):**
   \[
   y_n = f\left(\frac{1}{n+1}\right) + f\left(\frac{1}{n+2}\right) + \cdots + f\left(\frac{1}{2021n}\right)
   \]

2. **Rewrite \( y_n \) using \( x_n \):**
   \[
   y_n = \sum_{k=n+1}^{2021n} f\left(\frac{1}{k}\right)
   \]
   \[
   y_n = \sum_{k=1}^{2021n} f\left(\frac{1}{k}\right) - \sum_{k=1}^n f\left(\frac{1}{k}\right)
   \]
   \[
   y_n = x_{2021n} - x_n + \int_1^{2021n} f\left(\frac{1}{x}\right) \, dx - \int_1^n f\left(\frac{1}{x}\right) \, dx
   \]
   \[
   y_n = x_{2021n} - x_n + \int_n^{2021n} f\left(\frac{1}{x}\right) \, dx
   \]

3. **Evaluate the integral:**
   \[
   \int_n^{2021n} f\left(\frac{1}{x}\right) \, dx = \int_1^{2021} n f\left(\frac{1}{nu}\right) \, du
   \]
   Since \( f\left(\frac{1}{nu}\right) \sim \frac{1}{nu} \) for large \( n \):
   \[
   \int_1^{2021} n f\left(\frac{1}{nu}\right) \, du \sim \int_1^{2021} \frac{1}{u} \, du = \ln 2021
   \]

4. **Conclude the limit:**
   \[
   \lim_{n \to \infty} y_n = \lim_{n \to \infty} \left( x_{2021n} - x_n \right) + \ln 2021
   \]
   Since \( x_n \) converges to a constant, \( x_{2021n} - x_n \to 0 \):
   \[
   \lim_{n \to \infty} y_n = \ln 2021
   \]

The final answer is \( \boxed{ \ln 2021 } \)