Question

Let $0<\alpha<\beta<1$. Then $\lim\limits _{n \rightarrow \infty} \displaystyle\sum_{k=1}^{n} \int\limits_{\frac{1}{\left(k+\beta\right)}}^{\frac{1}{\left(k+\alpha\right)}} \frac{d x}{1+x}$ is

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

Answer 1

Answer: (B)

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