Question 344 of 401MediumMCQ
\[\mathop {\lim }\limits_{n \to \infty } {\left( {1 + {{1 + {1 \over 2} + ........ + {1 \over n}} \over {{n^2}}}} \right)^n}\] is equal to :
Solutions · 1
Official solution
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
\[ \therefore \] \[L = {e^{\mathop {\lim }\limits_{P \to \infty } {{(P + 1)} \over {{2^P}}}}}\]
\[ \Rightarrow L = {e^o} = 1\] Answer the question to unlock