Question

Evaluate $\underset{n\to \infty }{lim}[(1+\frac{1}{n})(1+\frac{2}{n})....(1+\frac{n}{n})]$

Answer 1

${lim}_{n\to \infty }\left[\right(1+\frac{1}{n}\left)\right(1+\frac{2}{n})\cdot \cdot \cdot (1+\frac{n}{n}\left)\right]$

Let $y={lim}_{n\to \infty }\left[\right(1+\frac{1}{n}\left)\right(1+\frac{2}{n})\cdot \cdot \cdot (1+\frac{n}{n}\left)\right]$

Taking log on both the sdes,

$\log y={lim}_{n\to \infty }\left[\log \right(1+\frac{1}{n}\left)\log \right(1+\frac{2}{n})\cdot \cdot \cdot \log (1+\frac{n}{n}\left)\right]$

$={lim}_{n\to \infty }{\sum }_{r=1}^n\log (1+\frac{r}{n})$

Now, let $\frac{r}{n}=x$

When, $r=1,x=\frac{1}{n}=\frac{1}{\alpha }=0$

When $r=n,x=1$

$\therefore \log y={\int }_0^1\left(\log \right(1+x)\cdot 1)dx$

$=\left[\log \right(1+x)x{]}_0^1-{\int }_0^3(\frac{(1+x)-1}{1+x}dx$

$=\left[\log \right(1+x)x{]}_0^1-[x-\log (1+x){]}_0^1$

$=\log 2-1+\log 2$

$=\log 4-\log 10({\log }_{10}10=1)$

$=\log \left(\frac{4}{10}\right)$

$\therefore y=\frac{2}{5}$