Question

For each positive integer $n$, let
$y_n=\frac{1}{n}\left\{\right(n+1\left)\right(n+2)......(n+n){\}}^{\frac{1}{n}}$

For $x\in R$ let $\left[x\right]$ be the greatest integer less than or equal to $x$. If $\underset{n\to \infty }{lim}{y}_n=L$, then the value of $\left[L\right]$ is

Answer 1

$y_n=\frac{1}{n}\left\{\right(n+1\left)\right(n+2)......(n+n){\}}^{\frac{1}{n}}$
This can be written as,
$y_n=\prod _{r=1}^n{(1+\frac{r}{n})}^{\frac{1}{n}}$
Taking log on both sides,
$\ln y_n=\sum _{r=1}^n\frac{1}{n}\ln (1+\frac{r}{n})\Rightarrow \underset{n\to \infty }{lim}\ln y_n=\underset{n\to \infty }{lim}\sum _{r=1}^n\frac{1}{n}\ln (1+\frac{r}{n})=\underset{0}{\overset{1}{\int }}\ln (1+x)dx$
using integration by parts
$\Rightarrow \underset{n\to \infty }{lim}\ln y_n={\left[\right(x+1\left)\right\{\ln (x+1)-x\left\}\right]}_0^1\Rightarrow \ln L=2\ln 2-1=\ln \frac{4}{e}\therefore L=\frac{4}{e}1<L<2$
Hence $\left[L\right]=1$

Alternate solution:
We can say that,
$y_n<\frac{1}{n}\left\{\right(n+n\left)\right(n+n)......(n+n){\}}^{\frac{1}{n}}\Rightarrow y_n<\frac{1}{n}(2n{)}^{\frac{n}{n}}\Rightarrow y_n<2$
Now,
$y_n>\frac{1}{n}\left\{\right(n\left)\right(n)......(n){\}}^{\frac{1}{n}}\Rightarrow y_n>\frac{1}{n}(n{)}^{\frac{n}{n}}\Rightarrow y_n>1$
$\therefore 1<\underset{n\to \infty }{lim}{y}_n<2\Rightarrow 1<L<2$
Hence $\left[L\right]=1$