Question

Evaluate $\underset{n\to \infty }{lim}\left\{\right(1+\frac{1}{n}\left)\right(1+\frac{2}{n})\dots (1+\frac{n}{n}){\}}^{1/n}$

• A. $e$
• B. $1/e$
• C. $4/e$
• D. $8/e$

Answer 1

The correct option is C $4/e$

$\underset{n\to \infty }{lim}{((1+\frac{1}{n})(1+\frac{2}{n})\dots .(1+\frac{n}{n}))}^{\frac{1}{n}}$

$\underset{n\to \infty }{lim}{e}^{ln{[(1+\frac{1}{n})(1+\frac{2}{n})\dots .(1+\frac{n}{n})]}^{\frac{1}{n}}}(\because a^{{\log }_{a}m}=m)$

$\underset{n\to \infty }{lim}{e}^{\frac{1}{n}}(ln(1+\frac{1}{n})+ln(1+\frac{2}{n})+\dots .+ln(1+\frac{n}{n}))(\because ln(ab)=lna+lnb)$

$=e^{\underset{n\to \infty }{lim}\frac{1}{n}\cdot \sum _{r=1}^{n}ln(1+\frac{r}{n})}$

$=e^{{\int }_0^{1}ln(1+x).dx}$

$\{\begin{array}{c}{\int }_0^{1}ln(1+x)dx=ln(1+x)\cdot x{]}_0^1-{\int }_0^1\frac{1}{x+1}\cdot x\cdot dx\end{array}$

$=ln2-{\int }_0^1\frac{x+1–x}{x+1}dx$

$=ln2-{\int }_0^1\left(x\right)dx+{\int }_0^1\frac{1}{x+1}dx$

$=ln2–1+ln(x+1){]}_0^1$

$\begin{array}{c}=2ln2–1=ln\left(4/e\right)\end{array}\}$

$=e^{ln\left(4/e\right)}=\frac{4}{e}$.