limn 0.05em 1n( 1e^1/n+ 1e^2/n+ 1e^3/n+ …+ 1e^2)=

Question

Accepted Answer

Correct answer: A. (a) lim_h 0.05em 1n( 1e^1/n+ 1e^2/n+ 1e^3/n+ …+ 1e^2) = lim_h 0.05em 1n _k=1^2n 0.05em 1e^k/n = _0^2 0.05em 1e^x dx= _0^2 0.05eme^-x dx=(-e^-x)_0^2 =-e^-2+e^0=1-e^-2 .

$lim n \to \infty \frac{1}{n} (\frac{1}{e ^{1/ n}} + \frac{1}{e ^{2/ n}} + \frac{1}{e ^{3/ n}} + \dots + \frac{1}{e ^{2}}) =$

Solution

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