Official solution
\[{U_n} = \prod\limits_{r = 1}^n {{{\left( {1 + {{{r^2}} \over {{n^2}}}} \right)}^r}} \]
\[L = \mathop {\lim }\limits_{n \to \infty } {({U_n})^{ - 4/{n^2}}}\]
\[\log L = \mathop {\lim }\limits_{n \to \infty } {{ - 4} \over {{n^2}}}\sum\limits_{r = 1}^n {\log {{\left( {1 + {{{r^2}} \over {{n^2}}}} \right)}^r}} \]
\[ \Rightarrow \log L = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n { - {{4r} \over n}.{1 \over n}\log \left( {1 + {{{r^2}} \over {{n^2}}}} \right)} \]
\[ \Rightarrow \log L = - 4\int\limits_0^1 {x\log (1 + {x^2})\,dx} \]
put 1 + x2 = t
Now, 2xdx = dt
\[ = - 2\int\limits_1^2 {\log (t)dt = - 2[t\log t - t]_1^2} \]
\[ \Rightarrow \log L = - 2(2\log 2 - 1)\]
\[\therefore\] \[L = {e^{ - 2(2\log 2 - 1)}}\]
\[ = {e^{ - 2\left( {\log \left( {{4 \over e}} \right)} \right)}}\]
\[ = {e^{\log {{\left( {{4 \over e}} \right)}^2}}}\]
\[ = {\left( {{e \over 4}} \right)^2} = {{{e^2}} \over {16}}\]