Question

If $f\left(n\right)={[(n+1)(n+2)...(n+n)]}^{1/n}$, then ${lim}_{n\to \infty }f\left(n\right)$ equals

• A. $e$
• B. $1/e$
• C. $2/e$
• D. $4/e$

Answer 1

The correct option is D $4/e$

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