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(B) The integral is given by $\int_{1/(k+\beta)}^{1/(k+\alpha)} \frac{dx}{1+x} = [\log(1+x)]_{1/(k+\beta)}^{1/(k+\alpha)} = \log(1+\frac{1}{k+\alpha})

Vedclass · Accepted Answer

$\log_{e} \frac{1+\beta}{1+\alpha}$

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(B) The integral is given by $\int_{1/(k+\beta)}^{1/(k+\alpha)} \frac{dx}{1+x} = [\log(1+x)]_{1/(k+\beta)}^{1/(k+\alpha)} = \log(1+\frac{1}{k+\alpha}) - \log(1+\frac{1}{k+\beta})$.Summing from $k=1$ to $n$,we get $\sum_{k=1}^{n} (\log(\frac{k+\alpha+1}{k+\alpha}) - \log(\frac{k+\beta+1}{k+\beta}))$.This is a telescoping sum: $(\log(\frac{2+\alpha}{1+\alpha}) - \log(\frac{2+\beta}{1+\beta})) + (\log(\frac{3+\alpha}{2+\alpha}) - \log(\frac{3+\beta}{2+\beta})) + \dots + (\log(\frac{n+\alpha+1}{n+\alpha}) - \log(\frac{n+\beta+1}{n+\beta}))$.As $n \rightarrow \infty$,the terms $\log(\frac{n+\alpha+1}{n+\alpha}) \rightarrow \log(1) = 0$ and $\log(\frac{n+\beta+1}{n+\beta}) \rightarrow \log(1) = 0$.The remaining terms are $\log(\frac{1+\beta}{1+\alpha})$.

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

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