Discussion Forum

For each positive integer n,  let $y_{n}=\frac{1}{n}((n+1)(n+2)...(n+n))^{\frac{1}{n}}. $. For x ε R, let [x] be the greatest integer less than or equal to x. If $ \lim_{n \rightarrow\infty} y_{n}=L$, then the value of [L] is ...........

Answer:

Explanation:

We have 

       $y_{n}=\frac{1}{n}((n+1)(n+2)...(n+n))^{\frac{1}{n}}$    and   

                                                                                                           $\lim_{n \rightarrow\infty}y_{n}=L$

   $\Rightarrow$    L=  $\lim_{n \rightarrow\infty}\frac{1}{n}[(n+1)(n+2)(n+3)....(n+n)]^{\frac{1}{n}}$

  $\Rightarrow$    L= $\lim_{n \rightarrow \infty}\frac{1}{n}[(1+\frac{1}{n})(1+\frac{2}{n})(1+\frac{3}{n})....(1+\frac{n}{n})]^{\frac{1}{n}}$

$\Rightarrow$    $\log_{}{L}=\lim_{n \rightarrow\infty}\frac{1}{n}[\log_{}{}(1+\frac{1}{n})+\log_{}{}(1+\frac{2}{n})+\log_{}{}(1+\frac{3}{n})....\log_{}{}(1+\frac{n}{n})]^{\frac{}{}}$

$\Rightarrow$     $\log_{}{L}= \lim_{n \rightarrow\infty}\frac{1}{n}\sum_r^n\log_{}{(1+\frac{r}{n})}$

  $\Rightarrow$  $\log_{}{L}= \int_{0}^{1}1 \times \log_{}{(1+x)dx} $

$\Rightarrow$   $\log_{}{L}= (x.\log_{}{(1+x)_0^1}-\int_{0}^{1}[\frac{\text{d}}{\text{d}x}(\log_{}{(1+x)\int_{}^{} dx]dx} $

                           [ by using integration by parts]

$\Rightarrow$   $\log_{}{L}= [x \log_{}{(1+x)_0^1}-\int_{0}^{1} \frac{x}{1+x}dx$

$\Rightarrow$     $\log_{}{L}= \log_{}{2}-\int_{0}^{1}(\frac{x+1}{x+1}-\frac{1}{x+1})dx $

 $\Rightarrow$   $\log_{}{L}= \log_{}{2}-[x]_0^1+[\log_{}{(x+1)]_0^1} $

$\Rightarrow$     $\log_{}{L}= \log_{}{2}-1+\log_{}{2}-0 $

$\Rightarrow$   $\log_{}{L}= \log_{}{4}-\log_{}{e}=\log_{}{\frac{4}{e}}$

$\Rightarrow$    $L=\frac{4}{e}$

$\Rightarrow$  [L]=[$\frac{4}{e}$]=1