Calculate lim_(n→+∞) A_n =∫


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Question Number 172307 by mathocean1 last updated on 25/Jun/22

$C a l c u l a t e$ $Double subscripts: use braces to clarify$
	Answered by aleks041103 last updated on 25/Jun/22

	$A_{n} = \int_{0}^{1} \frac{x^{n}}{1 - (- x)} d x = {(- 1)}^{n} \int_{0}^{1} \frac{{(- x)}^{n} - 1 + 1}{1 - (- x)} d x =$ $= {(- 1)}^{n + 1} \int_{0}^{1} \frac{1 - {(- x)}^{n} - 1}{1 - (- x)} d x =$ $= {(- 1)}^{n + 1} [\int_{0}^{1} \frac{1 - {(- x)}^{n}}{1 - (- x)} d x - \int_{0}^{1} \frac{d x}{1 + x}] =$ $= {(- 1)}^{n + 1} [\int_{0}^{1} (\underset{k = 1}{\sum^{n}} {(- x)}^{k - 1}) d x - l n (1 + 1) + l n (1 + 0)] =$ $= {(- 1)}^{n + 1} [\underset{k = 1}{\sum^{n}} {(- 1)}^{k - 1} (\int_{0}^{1} x^{k - 1} d x) - l n (2)] =$ $= {(- 1)}^{n + 1} [\underset{k = 1}{\sum^{n}} \frac{{(- 1)}^{k - 1}}{k} - l n (2)]$ $\frac{1}{1 + x} = \underset{k = 0}{\sum^{\infty}} {(- x)}^{k}$ $\Rightarrow \int_{0}^{s} \frac{d x}{1 + x} = l n (1 + s) = \underset{k = 0}{\sum^{\infty}} {(- 1)}^{k} \frac{s^{k + 1}}{k + 1} =$ $\Rightarrow l n (1 + s) = \underset{k = 1}{\sum^{\infty}} \frac{{(- 1)}^{k - 1} s^{k}}{k}$ $\Rightarrow \underset{k = 1}{\sum^{n}} \frac{{(- 1)}^{k - 1}}{k} = l n (2) - \underset{k = n + 1}{\sum^{\infty}} \frac{{(- 1)}^{k - 1}}{k}$ $\Rightarrow A_{n} = {(- 1)}^{n} \underset{k = n + 1}{\sum^{\infty}} \frac{{(- 1)}^{k - 1}}{k}$ $Double subscripts: use braces to clarify$
	Answered by Mathspace last updated on 25/Jun/22

	$A_{n} = \int_{R^{+}} \frac{x^{n}}{1 + x} χ_{[0, 1]} (x) d x$ $= \int_{R^{+}} f_{n} (x) d x$ $f_{m} c o n v e r g e s i m p l e m e n t v e r s 0$ $c a r x \in [0, 1] a n d f_{n} e s t d o m i n e e p a r$ $g (x) = \frac{1}{1 + x} \Rightarrow l i m A_{n} = \int_{R^{+}} l i m f_{n} = 0$
	Answered by puissant last updated on 25/Jun/22

	$0 ⩽ x ⩽ 1 \Rightarrow 1 ⩽ 1 + x ⩽ 2$ $o n a : \frac{x^{n}}{2} ⩽ \frac{x^{n}}{1 + x} ⩽ x^{n}$ $\Rightarrow \frac{1}{2} \int_{0}^{1} x^{n} d x ⩽ I_{n} ⩽ \int_{0}^{1} x^{n} d x$ $\Rightarrow \frac{1}{2 (n + 1)} ⩽ I_{n} ⩽ \frac{1}{n + 1}$ $d o n c \lim_{n \to + \infty} I_{n} = 0$ $d^{'} a p r e s l e t h e o r e m e d e s g e n d a r m e s . .$
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