Question

Let $0<\alpha <\beta <1$. Then $\underset{n\to \infty }{lim}\sum _{k=1}^n\underset{1/(k+\beta )}{\overset{1/(k+\alpha )}{\int }}\frac{dx}{1+x}$ is

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

Answer 1

The correct option is B ${\log }_e\frac{1+\beta }{1+\alpha }$

$\underset{n\to \infty }{lim}\sum _{k=1}^n\underset{1/(k+\beta )}{\overset{1/(k+\alpha )}{\int }}\frac{dx}{1+x}$
$=\underset{n\to \infty }{lim}\sum _{k=1}^n[\ln (1+x){]}_{1/(k+\beta )}^{1/(k+\alpha )}$
$=\underset{n\to \infty }{lim}\sum _{k=1}^n[\ln (1+\frac{1}{k+\alpha })-\ln (1+\frac{1}{k+\beta })]$
$=\underset{n\to \infty }{lim}\sum _{k=1}^n\ln \left[\left(\frac{k+\beta }{k+\alpha }\right)\left(\frac{k+\alpha +1}{k+\beta +1}\right)\right]$
$=\ln [\frac{\beta +1}{\alpha +1}\times \frac{\alpha +2}{\beta +2}\times \frac{\beta +2}{\alpha +2}\times \frac{\alpha +3}{\beta +3}\times \dots ]$
$={\log }_e\frac{1+\beta }{1+\alpha }$