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Question Number 137482 by Algoritm last updated on 03/Apr/21

	Answered by MJS_new last updated on 03/Apr/21

	$0$
	Commented by Algoritm last updated on 03/Apr/21

	$proof$
	Answered by TheSupreme last updated on 03/Apr/21

	$\int_{n}^{2 n} \frac{1}{2 x^{4}} ⩽ \int_{n}^{2 n} \frac{x d x}{1 + x^{5}} ⩽ \int_{n}^{2 n} \frac{1}{x^{4}} d x$ $f o r n > 1 w e h a v e x^{5} > 1 > 0$ $- \frac{1}{6} x^{- 3} ∣_{n}^{2 n} ⩽ I ⩽ - \frac{1}{3} x^{- 3} ∣_{n}^{2 n}$ $[- \frac{1}{48 n^{3}} + \frac{1}{6 n^{3}}] ⩽ I ⩽ [- \frac{1}{24 n^{3}} + \frac{1}{3 n^{2}}]$ $\frac{7}{48 n^{3}} ⩽ I_{n} ⩽ \frac{7}{24 n^{3}}$ $f o r n \to \infty$ $0 ⩽ I ⩽ 0 \to I = 0$
	Commented by mathmax by abdo last updated on 03/Apr/21
	$let U_n =∫_n ^(2n) ((xdx)/(1+x^5 )) ⇒U_n =_(x=(1/t)) ∫_(1/n) ^(2/n) (1/(t(1+(1/t^5 ))))(−(dt/t^2 )) =−∫_(1/n) ^(2/n) (dt/(t^3 +(1/t^2 ))) =−∫_(1/n) ^(2/n) (t^2 /(1+t^5 ))dt we have 1+t^5 >1 ⇒ ∫_(1/n) ^(2/n) (t^2 /(1+t^5 ))dt <∫_(1/n) ^(2/n) t^2 dt =[(1/3)t^3 ]_(1/n) ^(2/n) =(1/3){(8/n^3 )−(1/n^3 )}→0(n→+∞) ⇒ lim_(n→+∞) U_n =0$
	$let U_{n} = \int_{n}^{2 n} \frac{xdx}{1 + x^{5}} \Rightarrow U_{n} =_{x = \frac{1}{t}} \int_{\frac{1}{n}}^{\frac{2}{n}} \frac{1}{t (1 + \frac{1}{t^{5}})} (- \frac{dt}{t^{2}})$ $= - \int_{\frac{1}{n}}^{\frac{2}{n}} \frac{dt}{t^{3} + \frac{1}{t^{2}}} = - \int_{\frac{1}{n}}^{\frac{2}{n}} \frac{t^{2}}{1 + t^{5}} dt we have 1 + t^{5} > 1 \Rightarrow$ $\int_{\frac{1}{n}}^{\frac{2}{n}} \frac{t^{2}}{1 + t^{5}} dt < \int_{\frac{1}{n}}^{\frac{2}{n}} t^{2} dt = {[\frac{1}{3} t^{3}]}_{\frac{1}{n}}^{\frac{2}{n}} = \frac{1}{3} {\frac{8}{n^{3}} - \frac{1}{n^{3}}} \to 0 (n \to + \infty) \Rightarrow$ $\lim_{n \to + \infty} U_{n} = 0$
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