Question

The value of $\underset{n\to \infty }{lim}{[{\left(\frac{n+1}{n}\right)}^{\lambda }.{\left(\frac{n+2}{n}\right)}^{\lambda }.........]}^{1/n}$

• A. ${\left(\frac{4}{e}\right)}^{\lambda }$
• B. ${\left(\frac{1}{e}\right)}^{\lambda }$
• C. ${\left(\frac{3}{e}\right)}^{\lambda }$
• D. ${\left(\frac{2}{e}\right)}^{\lambda }$

Answer 1

The correct option is A ${\left(\frac{4}{e}\right)}^{\lambda }$

Let $y=\underset{n\to \infty }{lim}{[{\left(\frac{n+1}{n}\right)}^{\lambda }.{\left(\frac{n+2}{n}\right)}^{\lambda }.........]}^{1/n}$
$\Rightarrow \log y=\frac{1}{n}\underset{n\to \infty }{lim}[\log {\left(\frac{n+1}{n}\right)}^{\lambda }+\log {\left(\frac{n+2}{n}\right)}^{\lambda }+........]$
$=\frac{1}{n}\underset{n\to \infty }{lim}\sum _{r=1}^n\left[\log {\left(\frac{n+r}{n}\right)}^{\lambda }\right]=\frac{1}{n}\underset{n\to \infty }{lim}\sum _{r=1}^n\lambda \log (1+\frac{r}{n})=\lambda {\int }_0^1\log (1+x)dx$
$=\lambda .2[\log 2-\frac{1}{2}]=\lambda [\log 4-\log e]=\lambda \log \left(\frac{4}{e}\right)$
$\Rightarrow \log y=\log {\left(\frac{4}{e}\right)}^{\lambda }$
$\therefore y={\left(\frac{4}{e}\right)}^{\lambda }$