Question

If $\lambda ={lim}_{n\to \infty }\frac{1}{n+\alpha }+\frac{1}{n+2\alpha }+\frac{1}{n+3\alpha }+....+\frac{1}{n+n\alpha }$ then $e^{\lambda }$ is

• A. ${(\alpha !)}^{\alpha }$
• B. ${(1+\alpha !)}^{\alpha !}$
• C. ${(1+\alpha )}^{1/\alpha }$
• D. $\frac{1}{\alpha }\log (1+\alpha )$

Answer 1

The correct option is C ${(1+\alpha )}^{1/\alpha }$

$\lambda ={lim}_{n\to \infty }\frac{1}{n+\alpha }+\frac{1}{n+2\alpha }+\frac{1}{n+3\alpha }+....+\frac{1}{n+n\alpha }$
$={lim}_{n\to \infty }{\sum }_{r=1}^n\frac{1}{n+\alpha r}={lim}_{n\to \infty }{\sum }_{r=1}^n\frac{1}{n}\left(\frac{1}{1+\alpha \frac{r}{n}}\right)$
$={\int }_0^1\frac{dx}{1+\alpha x}=\frac{1}{\alpha }{\left|\log (1+\alpha x)\right|}_0^1=\frac{1}{\alpha }\log (1+\alpha )$
$\therefore e^{\lambda }={(1+\alpha )}^{1/\alpha }$