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Question Number 144107 by SOMEDAVONG last updated on 21/Jun/21

A=lim_(n→+∝) ((1+(2)^(1/7) +(3)^(1/7) +(4)^(1/7) +.....+(n)^(1/7) )/( (n^9 )^(1/7) )) =?

$$\mathrm{A}=\underset{\mathrm{n}\rightarrow+\propto} {\mathrm{lim}}\frac{\mathrm{1}+\sqrt[{\mathrm{7}}]{\mathrm{2}}+\sqrt[{\mathrm{7}}]{\mathrm{3}}+\sqrt[{\mathrm{7}}]{\mathrm{4}}+.....+\sqrt[{\mathrm{7}}]{\mathrm{n}}}{\:\sqrt[{\mathrm{7}}]{\mathrm{n}^{\mathrm{9}} }}\:=? \\ $$

Answered by Dwaipayan Shikari last updated on 21/Jun/21

A=lim_(n→∞) ((1+(2)^(1/7) +..+(n)^(1/7) )/( (n^8 )^(1/7) ))  =(1/n)Σ_(k=1) ^n ((k/n))^(1/7) =∫_0 ^1 (x)^(1/7)  dx=(7/8)

$${A}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}+\sqrt[{\mathrm{7}}]{\mathrm{2}}+..+\sqrt[{\mathrm{7}}]{{n}}}{\:\sqrt[{\mathrm{7}}]{{n}^{\mathrm{8}} }} \\ $$$$=\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\sqrt[{\mathrm{7}}]{\frac{{k}}{{n}}}=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt[{\mathrm{7}}]{{x}}\:{dx}=\frac{\mathrm{7}}{\mathrm{8}} \\ $$

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