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Question-107080




Question Number 107080 by Khalmohmmad last updated on 08/Aug/20
Answered by abdomsup last updated on 08/Aug/20
A_n =((n^n {(n/n^n )+(n^2 /n^n )+....+1})/(n^n {(1/n^n ) +((2/n))^(n ) +((3/n^ ))^n +....1}))  A_n =(((1/n^(n−1) )+(1/n^(n−2) )+....+1)/(((1/n))^n  +((2/n))^n +...+1)) ⇒  lim_(n→+∞)  A_n =1
$${A}_{{n}} =\frac{{n}^{{n}} \left\{\frac{{n}}{{n}^{{n}} }+\frac{{n}^{\mathrm{2}} }{{n}^{{n}} }+….+\mathrm{1}\right\}}{{n}^{{n}} \left\{\frac{\mathrm{1}}{{n}^{{n}} }\:+\left(\frac{\mathrm{2}}{{n}}\right)^{{n}\:} +\left(\frac{\mathrm{3}}{{n}^{} }\right)^{{n}} +….\mathrm{1}\right\}} \\ $$$${A}_{{n}} =\frac{\frac{\mathrm{1}}{{n}^{{n}−\mathrm{1}} }+\frac{\mathrm{1}}{{n}^{{n}−\mathrm{2}} }+….+\mathrm{1}}{\left(\frac{\mathrm{1}}{{n}}\right)^{{n}} \:+\left(\frac{\mathrm{2}}{{n}}\right)^{{n}} +…+\mathrm{1}}\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} =\mathrm{1} \\ $$

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