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