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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 :

JEE · Math · previous-year question

  1. A.$${{{e^2}} \over {16}}$$correct
  2. B.$${4 \over e}$$
  3. C.$${{16} \over {{e^2}}}$$
  4. D.$${4 \over {{e^2}}}$$

Answer

A. $${{{e^2}} \over {16}}$$

Explanation

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

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