Tham khảo câu trả lời từ Trợ lý ảo Lazi To solve the given problem, we need to calculate the ratio \(\frac{A}{B}\), where: \[ A = \frac{1}{2022} + \frac{2}{2021} + \frac{3}{2020} + \cdots + \frac{2021}{2} + \frac{2022}{1} \] \[ B = \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots + \frac{1}{2022} + \frac{1}{2023} \] First, let's rewrite \(A\) in a more convenient form: \[ A = \sum_{k=1}^{2022} \frac{k}{2023-k} \] Now, let's rewrite \(B\): \[ B = \sum_{k=2}^{2023} \frac{1}{k} \] To find \(\frac{A}{B}\), we need to understand the behavior of these sums. Notice that \(A\) is a sum of fractions where the numerator increases while the denominator decreases, and \(B\) is a harmonic series starting from \(\frac{1}{2}\) to \(\frac{1}{2023}\). To simplify the calculation, let's consider the behavior of these sums for large \(n\). For large \(n\), the sum \(A\) can be approximated by the integral: \[ A \approx \int_{1}^{2022} \frac{x}{2023-x} \, dx \] Similarly, the sum \(B\) can be approximated by the integral: \[ B \approx \int_{2}^{2023} \frac{1}{x} \, dx \] Let's compute these integrals: \[ \int_{1}^{2022} \frac{x}{2023-x} \, dx = \int_{1}^{2022} \left( \frac{2023}{2023-x} - 1 \right) dx \] \[ = 2023 \int_{1}^{2022} \frac{1}{2023-x} \, dx - \int_{1}^{2022} 1 \, dx \] Let \(u = 2023 - x\), then \(du = -dx\), and the limits change from \(x=1\) to \(u=2022\) and from \(x=2022\) to \(u=1\): \[ = 2023 \left[ -\ln |2023-x| \right]_{1}^{2022} - (2022 - 1) \] \[ = 2023 \left( -\ln 1 + \ln 2022 \right) - 2021 \] \[ = 2023 \ln 2022 - 2021 \] For \(B\): \[ \int_{2}^{2023} \frac{1}{x} \, dx = \left[ \ln x \right]_{2}^{2023} \] \[ = \ln 2023 - \ln 2 \] \[ = \ln \frac{2023}{2} \] Now, the ratio \(\frac{A}{B}\) is: \[ \frac{A}{B} \approx \frac{2023 \ln 2022 - 2021}{\ln \frac{2023}{2}} \] This is a complex expression, but for large \(n\), the logarithmic terms dominate, and the constants become less significant. Therefore, the ratio simplifies approximately to: \[ \frac{A}{B} \approx \frac{2023 \ln 2022}{\ln \frac{2023}{2}} \] Given the complexity of the exact calculation, this approximation provides a reasonable estimate of the ratio.