Question

The value of $\underset{n\to \infty }{lim}\prod _{r=1}^n{\left(\frac{n+r}{n}\right)}^{\frac{1}{n}}$ is

• A. $4e$
• B. $\frac{4}{e}$
• C. $3e$
• D. $\frac{3}{e}$

Answer 1

The correct option is B $\frac{4}{e}$

Let $P=\underset{n\to \infty }{lim}\prod _{r=1}^n{\left(\frac{n+r}{n}\right)}^{\frac{1}{n}}$
$\Rightarrow P=\underset{n\to \infty }{lim}{[\frac{n+1}{n}\cdot \frac{n+2}{n}\cdots \frac{n+n}{n}]}^{\frac{1}{n}}$
Taking $\ln$ on both sides, we have
$\ln P=\underset{n\to \infty }{lim}\frac{1}{n}[\ln \left(\frac{n+1}{n}\right)+\ln \left(\frac{n+2}{n}\right)+\cdots +\ln \left(\frac{n+n}{n}\right)]$
$=\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=1}^n\ln (1+\frac{r}{n})=\underset{0}{\overset{1}{\int }}\ln (1+x)dx$
Using By parts,
$\ln P=\left[x\ln \right(1+x)-x+\ln (1+x){]}_0^1\Rightarrow \ln P=\ln 4-1\Rightarrow \ln P=\ln \left(\frac{4}{e}\right)\therefore P=\frac{4}{e}$