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

It is ${1^\infty }$ form<br><br>$$L = {e^{\mathop {\lim }\limits_{n \to \infty } \left( {{{1 + {1 \over 2} + {1 \over 3} + ...{1 \over n}} \over n}} \right)}}$$<br><br>$$S = 1 + \left( {{1 \over 2} + {1 \over 3}} \right) + \left( {{1 \over 4} + {1 \over 5} + {1 \over 6} + {1 \over 7}} \right) + \left( {{1 \over 8} + ......... + {1 \over {15}}} \right)$$<br><br>$$S &lt; 1 + \left( {{1 \over 2} + {1 \over 2}} \right) + \left( {{1 \over 4} + {1 \over 4} + {1 \over 4} + {1 \over 4}} \right).......... + \underbrace {\left( {{1 \over {{2^P}}} + ............ + {1 \over {{2^P}}}} \right)}_{{2^P}times}$$<br><br>$S &lt; 1 + 1 + 1 + 1 + ....... + 1$<br><br>$S &lt; P + 1$<br><br>$\therefore$ $L = {e^{\mathop {\lim }\limits_{P \to \infty } {{(P + 1)} \over {{2^P}}}}}$<br><br>$\Rightarrow L = {e^o} = 1$