If $${U_n} = \left( {1 + {1 \over {{n^2}}}} \right)\left( {1 + {{{2^2}} \over {{n^2}}}} \right)^2.....\left( {1 + {{{n^2}} \over {{n^2}}}} \right)^n$$, then $\mathop {\lim }\limits_{n \to \infty } {({U_n})^{{{ - 4} \over {{n^2}}}}}$ is equal to :

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