JEE MAIN - Mathematics (2021 - 25th February Morning Shift - No. 9)

$$\mathop {\lim }\limits_{n \to \infty } {\left( {1 + {{1 + {1 \over 2} + ........ + {1 \over n}} \over {{n^2}}}} \right)^n}$$ is equal to :
$${{1 \over 2}}$$
1
0
$${{1 \over e}}$$

Explanation

It is $${1^\infty }$$ form

$$L = {e^{\mathop {\lim }\limits_{n \to \infty } \left( {{{1 + {1 \over 2} + {1 \over 3} + ...{1 \over n}} \over n}} \right)}}$$

$$S = 1 + \left( {{1 \over 2} + {1 \over 3}} \right) + \left( {{1 \over 4} + {1 \over 5} + {1 \over 6} + {1 \over 7}} \right) + \left( {{1 \over 8} + ......... + {1 \over {15}}} \right)$$

$$S < 1 + \left( {{1 \over 2} + {1 \over 2}} \right) + \left( {{1 \over 4} + {1 \over 4} + {1 \over 4} + {1 \over 4}} \right).......... + \underbrace {\left( {{1 \over {{2^P}}} + ............ + {1 \over {{2^P}}}} \right)}_{{2^P}times}$$

$$S < 1 + 1 + 1 + 1 + ....... + 1$$

$$S < P + 1$$

$$ \therefore $$ $$L = {e^{\mathop {\lim }\limits_{P \to \infty } {{(P + 1)} \over {{2^P}}}}}$$

$$ \Rightarrow L = {e^o} = 1$$

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