Question

If $L=\underset{n\to \infty }{\mathrm{Lim}}\frac{1}{n^4}\prod _{i=1}^{2n}{\left(n^2+i^2\right)}^{\frac{1}{n}}$, then $\ln L$ is equal to

• A. $\ln 2+\frac{\pi }{2}-2$
• B. $2{\tan }^{-1}2-4+2\ln 5$
• C. $2{\tan }^{-1}2+4+2\ln 5$
• D. $2\ln 5-4-2{\tan }^{-1}2$

Answer 1

Answer: (B)

We have $L=\underset{n\to \infty }{\mathrm{Lim}}\frac{1}{n^4}\prod _{i=1}^{2n}{\left(n^2+i^2\right)}^{\frac{1}{n}}$
$\Rightarrow \ln L=\underset{n\to \infty }{\mathrm{Lim}}\frac{1}{n}\sum _{i=1}^{2n}\ln \left(n^2+i^2\right)-4\ln n$
$\Rightarrow \ln L=\underset{n\to \infty }{\mathrm{Lim}}\frac{1}{n}\sum _{i=1}^{2n}\ln \left(n^2\left(1+\frac{i^2}{n^2}\right)\right)-4\ln n$
$\Rightarrow \ln L=\underset{n\to \infty }{\mathrm{Lim}}\frac{2\ln n}{n}\sum _{i=1}^{2n}1+\frac{1}{n}\sum _{i=1}^{2n}\ln \left(1+\frac{i^2}{n^2}\right)-4\ln n$
$\Rightarrow \ln L=\underset{n\to \infty }{\mathrm{Lim}}\frac{2\ln n}{n}(2n)+\frac{1}{n}\sum _{i=1}^{2n}\ln \left(1+\frac{i^2}{n^2}\right)-4\ln n$
$\Rightarrow \ln L=\underset{0}{\overset{2}{\int }}\ln \left(1+x^2\right)dx={x\ln \left(1+x^2\right)|}_0^2-\underset{0}{\overset{2}{\int }}\frac{2x^2}{1+x^2}dx$
$\text{ So, }\ln L=2\ln 5-2\left[2-{\tan }^{-1}2\right]=2{\tan }^{-1}2-4+2\ln 5$