If $\underset{n\to \infty }{\lim }\frac{1}{n}\sum _{k=1}^{n}k\ln \left(\frac{n^2+{(k-1)}^2}{n^2+k^2}\right)$ exists and is equal to L. Find the absolute value of [L].
[Note : [y] denotes greatest integer less than or equals to y.]

$\underset{n\to \infty }{\lim }\frac{1}{n}\sum _{k=1}^n\ln {\left(\frac{n^2+{(k-1)}^2}{n^2+k^2}\right)}^k$

$=\underset{n\to \infty }{\lim }\frac{1}{n}\left(\ln ({\left(\frac{n^2+0^2}{n^2+1^2}\right)}^1.{\left(\frac{n^2+1^2}{n^2+2^2}\right)}^2.{\left(\frac{n^2+2^2}{n^2+3^2}\right)}^3\mathrm{.....}{\left(\frac{n^2+{(n-1)}^2}{n^2+n^2}\right)}^n)\right)$

$=\underset{n\to \infty }{\lim }\left[\ln \left(\frac{n^2.(n^2+1^2).(n^2+2^2)\mathrm{...}{(n^2+n-1)}^2}{{\left(2n^2\right)}^n}\right)\right]$

$=\underset{n\to \infty }{\lim }\frac{1}{n}\left[\ln (\frac{1}{2}.(\frac{1}{2}+\frac{1^2}{2n^2}).(\frac{1}{2}+\frac{2^2}{2n^2})\mathrm{...}(\frac{1}{2}+\frac{{(n-1)}^2}{2n^2}))\right]$

$\underset{n\to \infty }{\lim }\frac{1}{n}\sum _{r=0}^{n-1}\ln (\frac{1}{2}+\frac{r^2}{2n^2})={\int }_0^1\underset{II}{\underbrace{1}}.\underset{I}{\underbrace{\ln \left(\frac{1+x^2}{2}\right)dx}}$

$\Rightarrow S ={\int }_0^1\underset{II}{\underbrace{1}}.\ln \underset{I}{\underbrace{\left(\frac{1+x^2}{2}\right)dx}= x \ln {\left(\frac{1+x^2}{2}\right)|}_0^1-{\int }_0^1\frac{2x^2}{1+x^2}dx}$

$= 0 – 2[{\int }_0^1\frac{x^2+1}{1+x^2}dx-{\int }_0^1\frac{dx}{1+x^2}]$

$= –2[x–ta{n}^{–1} x{]}_0^1= –2 [1-\frac{\pi }{4}]=\frac{\pi }{2}– 2$

$$\begin{gathered} L=\frac{\mathrm{\pi }}{2}-2\Rightarrow \left[L\right]=-1 \\ \therefore \left|\left[L\right]\right|=1 \end{gathered}$$