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

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

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kritika · Answer 1 · 2021-12-26T19:32:33+0000

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

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

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