Q102 / 155 · WBJEE 2020 · Mathematics

Let $0 < \alpha < \beta < 1$. Then $\lim {n \rightarrow \infty} \sum_{k=1}^{n} \int_{1/k+\beta}^{1 /(k+a)} \frac{d x}{1+x}$ is

  • A
    $\log _{e} \frac{\beta}{\alpha}$
  • Correct answer
    B
    $\log _{e} \frac{1+\beta}{1+\alpha}$
  • C
    $\log _{e} \frac{1+\alpha}{1+\beta}$
  • D
    $\infty$

Solution

Hint: $\lim _{n \rightarrow \infty} \sum_{k=1}^{n}[\log |1+x|]_{\frac{1}{k+\beta}} \frac{1}{k+a}$ $=\lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(\log \left(1+\frac{1}{k+\alpha}\right)-\log \left(1+\frac{1}{k+\beta}\right)\right)$ $=\lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(\log \left(\frac{k+\alpha+1}{k+\alpha}\right)-\log \left(\frac{k+\beta+1}{k+\beta}\right)\right)$ $=\log \left(\frac{\beta+1}{\alpha+1}\right)$

Hence, the correct answer is option: 'B'.

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