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Q.
If $L=\underset {n \rightarrow \infty}{\text{Lim}} \frac{1}{n^4} \displaystyle\prod_{i=1}^{2 n}\left(n^2+i^2\right)^{\frac{1}{n}}$, then $\ln L$ is equal to
Integrals
Solution:
We have $L =\underset{ n \rightarrow \infty}{\text{Lim}} \frac{1}{ n ^4} \displaystyle\prod_{ i =1}^{2 n }\left( n ^2+ i ^2\right)^{\frac{1}{ n }}$
$\Rightarrow \ln L =\underset{ n \rightarrow \infty}{\text{Lim}} \frac{1}{ n } \displaystyle\sum_{ i =1}^{2 n } \ln \left( n ^2+ i ^2\right)-4 \ln n $
$\Rightarrow \ln L =\underset{ n \rightarrow \infty}{\text{Lim}}\frac{1}{ n } \displaystyle\sum_{ i =1}^{2 n } \ln \left( n ^2\left(1+\frac{ i ^2}{ n ^2}\right)\right)-4 \ln n $
$\Rightarrow \ln L =\underset{ n \rightarrow \infty}{\text{Lim}} \frac{2 \ln n }{ n } \displaystyle\sum_{ i =1}^{2 n } 1+\frac{1}{ n } \sum_{ i =1}^{2 n } \ln \left(1+\frac{ i ^2}{ n ^2}\right)-4 \ln n $
$\Rightarrow \ln L =\underset{ n \rightarrow \infty}{\text{Lim}}\frac{2 \ln n }{ n }(2 n )+\frac{1}{ n } \displaystyle\sum_{ i =1}^{2 n } \ln \left(1+\frac{ i ^2}{ n ^2}\right)-4 \ln n $
$\Rightarrow \ln L =\int\limits_0^2 \ln \left(1+ x ^2\right) dx =\left. x \ln \left(1+ x ^2\right)\right|_0 ^2-\int\limits_0^2 \frac{2 x ^2}{1+ x ^2} dx $
$\text { So, } \ln L =2 \ln 5-2\left[2-\tan ^{-1} 2\right]=2 \tan ^{-1} 2-4+2 \ln 5$