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WBJEE · Maths · Limits

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

  1. A \(\log _{e} \frac{\beta}{\alpha}\)
  2. B \(\log _{e} \frac{1+\beta}{1+\alpha}\)
  3. C \(\log _{e} \frac{1+\alpha}{1+\beta}\)
  4. D \(\infty\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\log _{e} \frac{1+\beta}{1+\alpha}\)

Step-by-step Solution

Detailed explanation

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