limn 0.05em(1+ 1+ 12+ … …+ 1nn^2)^n is equal to

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Accepted Answer

Correct answer: C. C JEE Main 2021 It is 1^ form L=e^ limz 0.05em ( 1+ 12+ 13+ … …+ 1nn ) S=1+( 12+ 13)+( 14+ 15+ 16+ 17)+( 18+ … … …+ 115) S<1+( 12+ 12)+( 14+ 14+ 14+ 14) … … …+( 12^P+ … … …+ 12^P)2^ Ptimes ) S<1+1+1+1+ … … …+1 S ∴m=e^ limn 0.05em (P+1 )2^P ⇒L=e^ =1

$lim n \to \infty (1 + \frac{1 + \frac{1}{2} + \dots \dots + \frac{1}{n}}{n ^{2}})^{n} is equal to$

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